Solving system of two linear equations

Question

Which method would you choose to solve the following system of equations? Solve it.

y=2x+1 (1)

3x+y=7 ​ (2)

Answer 1

Given the equations:

y=2x+1

3x+y=7

Since one equation is already in isolated variable format, we know that we will be using the substitution method to solve. In other words, taking one equation (with an isolated variable) and plugging it in for that variable into the other equation

So in this case, we will take our isolated y equation and use it for y in the second equation.

3x + (2x + 1) = 7 Combine like terms

5x + 1 = 7 Simplify and isolate the variable

5x + 1 - 1 = 7 - 1

5x = 6

5x / 5 = 6 / 5

x = 6/5 = 1.2

Now we will take this new value for x and plug it into the first equation.

y = 2(1.2) +1

y = 2.4 + 1

y = 3.4 = 17/5

So we have the answer of x = 1.2, y = 3.4. This is also the point of intersection of the two lines when graphed. (1.2, 3.4)

Answer 2

y is already defined in terms of x as 2x+1.

Plug (1) into (2) to get:

3x +(2x +1) =7

5x+1=7

5x = 6

x = 6/5

then put this answer into y:

2x+1 = 2*6/5 +1 = 12/5 +1 = 12/5 +5/5 =17/5 or 3 2/5

Answer 3

Rewrite the system as:

2x - y = -1

3x + y = 7

Add the equations to get 5x = 6.

So, x = 6/5 = 1.2.

Substitute into either of the original equations: 3(1.2) + y = 7

So, y = 7 - 3.6 = 3.4.

The two lines intersect at the point (1.2, 3..4).

Answer 4

1) Choose a method (graphing, substitution, elimination)

This system is perfect for the substitution method, because the equation (1) already has y isolated. Therefore, we'll substitute the expression for y into equation (2): 3x + 2x + 1 = 7

2) Simplify (combine like terms) and solve for x:

5x+1=7

What's our next step? 5x+6, x=6/5 or 1.2

3) Plug in 6/5 instead of x in (1) equation (we chose the first one because y is already on its own, so it will save time): y = 2(6/5) + 1= 12/5 + 1 = 12/5 +5/5= 17/5 or 3.4

Answer: x=6/5, y=17/5

Answer 5

Equations:

y=2x+1 ------à (1)

3x+y=7  ------à (2)

Solving these equations with the following steps.

Step 1: Analyze the equations with the given information:

Two equations with the same variables: x and y

A variable already alone on one side, i.e. Isolated variable – y

y = 2x +1

Step 2:

According to the given information:

We can plug in the value of ‘y ’ in equation 2.

3x+(2x+1)=7

Simplifying the above equation:

3x+2x+1 = 7

Step 3: Combine like terms

5x+1=7

Step 4: Solve for x

Subtract 1 from both sides:

5x=6

Divide by 5 both sides:

5x/5 = 6/5

X = 6/5

x=1.2

Step 5: Back‑substitute to find y

Now plug in the value of x in step 4 into equation 1

y=2x+1

y=2(1.2)+1

y=2.4+1

y=3.4

Final Answer:

(x,y)= (1.2, 3.4)

Answer 6

The solution to the system of equations is

x=1.2bold x equals 1.2

𝐱=𝟏.𝟐

and y=3.4bold y equals 3.4

𝐲=𝟑.𝟒

. 

➡️ Step 1: Selection of Method 

The Substitution Method is the most efficient choice because the first equation is already explicitly solved for yy

𝑦

. This allows for direct substitution into the second equation, avoiding the need for rearranging terms or using the elimination method. 

➡️ Step 2: Substitution and solving for xx

𝑥

 

Substitute the expression for yy

𝑦

from equation (1) into equation (2):

3x+(2x+1)=73 x plus open paren 2 x plus 1 close paren equals 7

3𝑥+(2𝑥+1)=7

Combine the like terms on the left side:

5x+1=75 x plus 1 equals 7

5𝑥+1=7

Subtract 1 from both sides:

5x=65 x equals 6

5𝑥=6

Divide by 5 to isolate xx

𝑥

:

x=1.2x equals 1.2

𝑥=1.2

➡️ Step 3: Solving for yy

𝑦

 

Substitute the calculated value of xx

𝑥

back into equation (1) to determine the value of yy

𝑦

:

y=2(1.2)+1y equals 2 open paren 1.2 close paren plus 1

𝑦=2(1.2)+1

y=2.4+1y equals 2.4 plus 1

𝑦=2.4+1

y=3.4y equals 3.4

𝑦=3.4

✅ Answer: 

The solution to the system of equations is x=1.2bold x equals 1.2

𝐱=𝟏.𝟐

and y=3.4bold y equals 3.4

𝐲=𝟑.𝟒

. Expressed as fractions, the solution is x=65bold x equals six-fifths

𝐱=𝟔𝟓

and y=175bold y equals seventeen-fifths

𝐲=𝟏𝟕𝟓

Answer 7

As y is already defined in terms of x for the first equation, we would want to substitute in for y in the second equation to get our answer.

y = 2x + 1

3x + y = 7

3x + 2x + 1 = 7

5x + 1 = 7

5x = 6

x = 6/5

From there, we can plug back into the first equation to figure out what y actually equals.

y = 2(6/5) + 1

y= 12/5 + 1

And we can simplify from there to get

y = 17/5

You can always plug these numerical values back into the original equations to check your answer.

Answer 8

It would be quicker and more practical to substitute equation (1) y =2x+1 into equation (2) 3x+y=7 by way of direct substitution.

(1) y =2x+1

(2) 3x+y=7

3x + y = 7

(2x+1) because y=2x+1 -----Same y coordinate

3x + 2x +1 = 7 add and subtract like terms

'Then 5x+1=7, 5x=6----->x=6/5 Solve for x

Now substitute the x value you just found x=6/5 into either of given equations above (1) or (2)

equation (1) y=2(6/5) +1 ----------> y=(12/5)+1

Now rationalize using the method of equivalent fractions---> y=(12/5)+(5/5)

--->y=17/5

The solution to the given system is the coordinate of intersection (6/5,17/5)

Answer 9

We have three options here to solve this system of equations. In the age of technology, we could solve by graphing using the Desmos calculator. I know in the state where I teach (PA), the Desmos calculator is embedded in our statewide Algebra I assessment.

Option 1: Graphing via the Desmos (desmos.com/calculator)

The beauty of the Desmos calculator is that you can type your equations into the calculator in any form: slope intercept, standard, point-slope form, etc. WIth a TI-84, all equations would have to be converted into slope-intercept form (y = mx+b).

When we solve systems of linear equations, we are looking for ordered pairs in which the two lines have in common, or points of intersection. Once graphed in Desmos, we can hover our mouse to where the two lines intersect, and clearly see that the solution we are looking for is (1.2, 3.4)

This method takes about 20 seconds from start to finish.

Option 2: Solve using substitution

This is more of a traditional way to solve this system (because we won't always have access to a computer to graph the lines). Substitution is a method where we replace a variable with an equivalent expression (sort of like replacing a $1 bill with 4 quarters).

Substitution makes sense here because one of our 4 variables is already isolated ( y = 2x+1) meaning we know that y is equivalent to 2x +1, hence we can SUBSTITUTE y our for 2x +1.

That is what we will do here. Let's take our second equation ( 3x + y = 7) and substitute 2x+1 in place of y. This will create an equation with only 1 variable which is then solvable.

3x + y = 7

SUBSTITUTE FOR Y

3x + 2x + 1 = 7

COMBINE LIKE TERMS

5x + 1 = 7

SUBTRACT BOTH SIDES BY 1

5x = 6

DIVIDE BOTH SIDES BY 6

x = 6/5 or x = 1.2

Because this is a terminating decimal, I'm okay with using 1.2 as my x. If it were non-terminating, we'd want to continue to use the fractional answer as our x since that is an exact answer.

Now we know that these two lines intersect at (1.2 , ? ), so we have to know use the x-value we found to figure out what the y-value is. The beauty here is that since these lines share this point, we can substitute x = 1.2 into either equation and still get the correct answer. I always tell my students, substitute into the equation that will require less work/be easier. For me, I'll substitute into y = 2x+1 since y is already isolated and that's my goal, figure out y.

y = 2x+ 1

SUBSTITUTE

y = 2(1.2) +1

SIMPLIFY USING ORDER OF OPERATIONS (multiply first, then add)

y = 3.4

So we've arrived at our answer. These two linear equations intersect at the point (1.2, 3.4). This is the only ordered pair that satisfies both equations.

This method takes about 2 minutes for a HS student from start to finish.

Option 3: Solve using Elimination

I am not going to show this method, because it is unnecessary and the question asked "which method would you choose?" When a variable is isolated to begin, substitution is usually the answer.

Answer 10

y = 2x + 1

3x + y = 7

The goal is to define the variables with numerical values, and with this system we are already given the value of y in terms of x, so substitution is simplest for most students.

Plug in our y value as 2x + 1 and solve for x:

3x + (2x + 1) = 7

3x + 2x + 1 = 7

5x + 1 = 7

5x = 6

x = 6/5 = 1 ¹/₅ = 1.2

.....which we can plug back into the given equations and solve for y to get a numerical value for it:

y = 2(6/5) + 1

= 12/5 + 1

= 12/5 + 5/5

= 17/5 = 3 ²/₅ = 3.4

Other students may quickly see that since y has no coefficients in either equation, it can be easily isolated & eliminated by rearranging terms and adding the equations together:

y = 2x + 1

3x + y = 7

2x - y = -1

3x + y = 7

5x = 6

x = 6/5

And plug our x value into one of the system's equations (#1 is easier, since it's already solved for y):

y = 2(6/5) + 1

= (12/5) + 1

= (12/5) + (5/5)

= 17/5

Answer 11

y=2x+1

3x+y=7

3x+(2x+1)=7

5x+1=7

5x=6

x=6/5

y=2(6/5)+1

y=12/5 + 1

y=12/5 + 5/5

y=17/5

Answer 12

Hello,

One method you can use is the elimination method. This method lines up the equations in a column, and manipulates the equations, either by multiplying or dividing, so that the coefficients, when added, equal 0.

3x+y=7

y=2x+1; Can be rewritten as 2x-y=-1

3x+y=7......(1)

2x-y=-1......(2)

For the y's, if you add (1)+(2) immediately, the y's cancel out to equal 0, leaving you with:

(3x+2x)+(y-y)=(7-1)

simplifying:

5x=6

therefore, x=6/5.

Now, you can plug this back into either equation to determine y.

Plug back into (1): 3(6/5)+y=7

Isolate y:

y=7-3(6/5)

simplify:

y=17/5

Plug back into (2): 2(6/5)-y=-1

isolate y:

y=1+2(6/5)

simplify:

y=17/5

Thus, y is definitely 17/5 as this process works when you plug in the x value for both equation (1) & (2).

A final check, plug in both the determined x and y values into equations (1) & (2):

3(6/5)+17/5=7

18/5+17/5=7

35/5=7

7=7; Correct!

2(6/5)-17/5=-1

12/5-17/5=-1

-5/5=-1

-1=-1; Correct!

In summary, x=-6/5, y=17/5

Hope this helps!

-Andrew Evans

Answer 13

y = 2x + 1 ----- (1)

3x + y = 7 ​ ----- (2)

Before we start, let me rearrange (2) so it's in the same format as (1), just for simplicity:

3x + y = 7

=> y = -3x + 7 ----- (3)

There are 3 ways to solve a system of linear equations:

1) GRAPHING METHOD:

Draw the line graphs of both linear equations. The point at which they intersect is your answer for both x and y values which work for both equations.

2) ELIMINATION METHOD:

Multiply (1) by a negative sign so it becomes:

- (y = 2x + 1)

=> -y = -2x - 1 ----- (4)

Do you see why I multiplied (1) by a negative sign? This helps us cancel out the y values when we add both equations, hence known as the Elimination Method. So now our simplified equation contains only 1 unknown variable x. And this is something we can easily solve. Adding (4) and (3) gives us:

-y = -2x - 1

y = -3x + 7

(the y values cancel out)

=> 0 = -5x + 6

=> 5x = 6

=> x = 6/5

Now insert this value of x into any of the original equations to get the value of y. Let's use (1):

y = 2x + 1

=> y = 2(6/5) + 1

=> y = 12/5 + 1

=> y = 17/5

3) SUBSTITUTION METHOD:

For this method, we will simply insert the value of y from equation 1 into 2. Using (2):

3x + y =7

(inserting the value of y from (1) )

3x + (2x + 1) = 7

(now again, we only have an equation with only 1 unknown variable, so we can easily solve it)

=> 5x + 1 = 7

=> 5x = 6

=> x = 6/5

Insert this value of x into any of the original equations to get the value of y, just like last time!

Happy Hunting!

Answer 14

Use a calculator like Desmos.

Type in the equations, check the point of intersection: (1.2, 3.4)

Answer 15

y = 2x + 1 (1)

3x + y = 7 ​ (2)

If this system of linear equations looks like this, then I would use substitution method because you can plug in one equation to the other to solve for one of the variables.

3x + (2x + 1) = 7

3x + 2x + 1 = 7

5x + 1 = 7

5x = 6

x = 6/5 or 1.2

Then substitute 6/5 for x in the original equation to solve for y.

y = 2(6/5) + 1

y = 12/5 + 1

y = 12/5 + 5/5

y = 17/5 or 3.4

Let's check our solution.

2(6/5) + 1 = 12/5 + 1 = 17/5

3(6/5) + 17/5 = 18/5 + 17/5 = 35/5 = 7

The solution to this system is (6/5, 17/5).

Answer 16

I would use the elimination method here! This requires aligning the y terms on one side of the equation and the x terms on the other. Equation 1 has the y terms on the left side of the equation and the x term on the right side. I will re-write equation 2 as follows in order to match that:

y=2x+1 (1) (given)

+3x-3x+y=7-3x ​ (2) (re-written by subtracting 3x from both sides)

I would then simply equation 2 by cancelling out the +3x and -3x terms:

y=7-3x (2)

Now we have the following equations:

y=2x+1 (1)

y=7-3x (2)

We now have to decide which variable to eliminate. We can multiply equation 2 by negative 1 to eliminate the y term from the left side of the equation as follows:

y=2x+1 (1)

-y=-7+3x (2; equation multiplied by negative 1)

We can then add the equations together as follows:

y-y=2x+1-7+3x

We can then simplify like terms:

0=5x-6

We can then add 6 to both sides of the equation and divide both sides by 5

0+6 = 5x -6 + 6 simplifies to

6 = 5x ---> then divide both sides by 5 to get

6/5 = 5x/5 which means

6/5 = x

We can then plug x= 6/5 back into either equation to solve for y. I will use equation 1

y=2x+1 where x= 6/5

y = 2 (6/5) + 1

y= 12/5 + 1 ---> rewrite 1 as 5/5

y = 12/5 + 5/5 = 17/5

The solution to the system is therefore x=6/5 and y=17/5.

To check your work, you can plug the x and y coordinate into both equations:

y=2x+1

17/5 = 2 *(6/5) + 1

17/5 = 12/5 + 5/5

17/5 = 17/5 ---> the solution is verified with equation 1

3x+y=7 ​

3 * (6/5) + 17/5 = 7

18/5 + 17/5 = 7

35/5 = 7

7 = 7 --->the solution is also verified with equation 2

Answer 17

Anytime one of the equations is already written in terms of x or in terms of y, substitution is the simplest way to solve the system. The first equation is written in terms of y, so in the second equation we are going to substitute 2x + 1 for y:

3x + 2x + 1 = 7

Next, combine like terms:

5x + 1 = 7

Next, subtract 1 from both sides of the equation:

5x +1 -1 = 7 - 1

5x = 6

Next, divide both sides of the equation by 5:

5/5 x = 6/5

x = 6/5

Now that we have found the value of x, we can find the value of y by substituting 6/5 in for x in the first equation:

y=2x+1

y = 2(6/5) + 1

y = 12/5 + 1

Rewrite 1 as 5/5 so you can add the fractions together:

y = 12/5 + 5/5 

y = 17/5

Finally, write your answer as an ordered pair:

(6/5, 17/5)

Answer 18

Find value of y in terms of x as 2x+1.

Put equation (1) into (2) to get:

3x +(2x +1) =7

5x+1=7

5x = 6

x = 6/5

then substitute the value of x in y to get y

2x+1 = 2*6/5 +1 = 12/5 +1 = 12/5 +5/5 =17/5

Answer 19

This problem can be solved in 2 different ways: substitution and elimination. Since one of the variables has the number 1 as its coefficient, both methods are equal in difficulty for this particular problem.

Using substition, let's rewrite just the second equation:

3x + y = 7 ​ Subtract 3x from both sides of the equation to create y = 7 - 3x

Our two equations are now:

y = 2x + 1

y = 7 - 3x

Both equations are now solved for the variable y. and can therefore be set equal to each other:

2x + 1 = 7 - 3x

Let's arrange like terms on separate side of the equal sign. Add 3x and subtract 1 to both sides of the equal sign to create:

2x + 3x = 7 - 1

This simplifies to:

5x = 6

Dividing both sides by 5 gives us:

x = 6/5

Now to find y, we choose either one of our initial equations, and replace x by the answer we just found, placed within parentheses. Let's choose the first equation:

y = 2(6/5) + 1

Distribute the 2:

y = 12/5 + 1

Rewrite 1 as 5/5, so that the fractions can be added together:

y = 12/5 + 5/5

y = 17/5

Therefore our pair is (6/5, 17/5)

Now let's try elimination. It works best if we can rearrange our equations so that of the variables in one equation is the opposite sign of the other. Let me show it. Rewrite just the 1st equation:

y=2x+1 Subtract y and subtract 1 to both sides: -1 = 2x -y

This is nearly ready, but elimination will works a little better if we rearrange our terms. Flip both sides around:

2x - y = -1

Now our two equations are this:

2x - y = -1

3x + y = 7

Notice how the x-terms, y-terms, equal sign, and constant are directly above and below each other. This is the setup we want. Elimination wants us to add these columns in such a way so that one variable will cancel out. Fortunately we can add the two lines directly and the y will disappear:

2x + 3x -> 5x

-y + y -> 0

-1 + 7 -> 6

The y-value has become 0, meaning we have successfully eliminated it. Our new equation is:

5x = 6

Which is the same equation we got before during the substitution method, meaning x will again be:

x = 6/5

As for the y-value, we solve it the same way as we did using substitution. Our pair is (6/5, 17/5)

It is more time efficient to use elimination only when solving one of the two variables. Less steps will be required because substituting a number into one equation gets us to our answer faster, rather than trying to eliminate the x variable using 2 equations.

Answer 20

Given System is

y=2x+1 -----------(1)

3x+y=7 -----------(2)

we rearrange the system as

2x-y=-1 ------------(3)

3x+y=7 ------------(4)

This system we can solve by elimination or substitution method ( Because eq (1) already express y in terms of x)

Step 1: Substitute equ (1) y=2x+1 in equ (2)

3x + y = 7

3x + (2x+1) = 7

Step 2: Simplify

3x + 2x + 1 = 7

5x = 7-1

5x = 6

Step 3: Solve for x x = 6/5

Step 4: Substitute x=6/5 back into eq(1)

y = 2x + 1

y = 2 (6/5) +1

Step 5: Simplify

y = 12/5 +1

y = 12/5 +5/5 (By taking L.C.M)

y = (12+5) /5

y = 17 / 5

Final Answer is x = 6/5 and y = 17/5

Answer 21

Which method would you choose to solve the following system of equations? Solve it.

y=2x+1 (1)

3x+y=7 ​ (2)

I would use the substitution method. The first equation is y = 2x+1 so we can plug 2x +1 in for y on the second equation.

3x+(2x + 1) = 7

Now we can combine like terms so 3x + 2x = 5x

5x + 1 = 7

Now we can begin to solve (get x by itself) by subtracting 1 from both sides.

5x = 6

Next, we divide 5 from both sides so the 5 cancels out on the left side leaving x by itself.

x = 6/5 or 1 1/5

Now we need to solve for y so we will plug 6/5 in for x using the y = 2x +1 equation

y = 2 * 6/5 + 1

2*6/5 = 12/5 +1 = 17/5

y=3 2/5

Answer 22

Hi there. There are three ways to do this problem, either by elimination, substitution, or graphing. The easiest though is by substitution because that method is preferred if one equation is already solved for a variable, and our first equation is.

1) Label each equation (1) and (2) so you can label your work as you go - prevents getting lost!

(1) y = 2x + 1

(2) 3x + y = 7

2) You can plug (1) into (2) or (2) into (1). It does not matter. However, it will be easier if you plug (1) into (2) because in (1), the "y" is already alone.

(1) into (2): 3x + (y) = 7 ==> 3x + (2x + 1) = 7 (this was the "substitution" step) - the goal is to get rid of

one variable, which you did! Congratulations!)

Simplify (2): 3x + 2x + 1 = 7 (drop parentheses because they're not needed)

5x + 1 = 7 (combine like terms)

5x + 1 = 7 (isolate the "x" by subtracting 1 from both sides, the inverse operation of addition)

=============

5x = 6 (Isolate the "x" more by dividing both sides by 5, the inverse operation of multiplication)

x = 6/5

Solve for y:

Plug x = 6/5 into (1): y = 2x + 1 ==> y = 2 (6/5) + 1

y = 12/5 + 1 (multiplied 2(6/5) = 2/1 x 6/5 = (2 x 6) / (1 x 5) = 12/5)

y = 12/5 + 5/5 (common denominator 1 = 1/1 = 5/5)

y = 17/5 (to add fractions with like denominators, keep same denominator

and add the numerators)

So, the solution is {(x,y)} = {(6/5, 17/5)} (Recall, to write the answer of a system, write your x and y in an ordered pair (x, y), and then put that into the braces { } like this: { ( x, y ) }

Check: You can check your solution in either equation, and they should both give you a true equality.

(1) y = 2x + 1 ==> 17/5 = 2(6/5) + 1 ==> 17/5 = 12/5 + 5/5 ==> 17/5 = 17/5 Check!

(2) 3x + y = 7 ==> 3(6/5) + (17/5) = 7 ==> 18/5 + 17/5 = 7 ==> 35/5 = 7 ==> 7 = 7 Check!

Answer 23

2x +1 = y

3x +y=7

y = 7-3x = 2x+1

5x = 6

x = 6/5 = 1.2

y =3.4

use either substitution elimination, matrix algebra with row operations or graph the 2 equations and find their intersection point (x,y) = (1.2, 3.4)

Answer 24

I have explained the solution through the attached video.

https://youtu.be/BwMZM7oMD54

Answer 25

I would recommend substitution, because equation (1) is already solved for y:

First, substitute y = 2x + 1 into equation (2):

3x + (2x + 1) = 7

5x + 1 = 7

5x = 6

x = 6/5

Next, substitute x back into equation (1):

y = 2 * (6/5) + 1

y = 12/5 + 5/5

y = 17/5

Therefore, the solution is:

x = 6/5

y = 17/5

Answer 26

To solve this system of equations, the most straightforward method would be to use the substitution method since we already have y in terms of x , (y=2x+1).

therefore,

3x+y=7

Step 1: Substitute, 3x+(2x+1)=7

3x +2x+1=7

Step 2: Combine like terms, 5x+1=7

Step 3: Solve the two step equation for x,

5x=6

x=6/5

OR another method,

y=2x+1

3x+y=7

using the second equation, rearrange to get,

y=7-3x

now, y=2x+1 and y= 7-3x

therefore,

2x+1=7-3x

combine like terms, 5x= 6

x=6/5

Answer 27

Since equation 1 is already solved for y, I would use the substitution method.

Step 1: Solve for x by substituting the value of y in equation 1 (2x + 1) into y of equation 2

--> 3x + 2x + 1 = 7

--> 5x + 1 = 7

--> 5x = 6

--> x = 6/5

Step 2: Solve for y by substituting x = 6/5 into equation 1

--> y = 2(6/5) + 1

--> y = 12/5 + 1

--> y = 12/5 + 5/5

--> y = 17/5

Final Answer: x = 6/5, y = 17/5

Answer 28

WAY 1:

Line up the like terms by moving the 2x to the left side (Subtract 2x from each side of equation 1):

-2x + y = 1 (1)

3x + y = 7 ​ (2)

Subtract (2) from (1) to get this:

-5x = -6

Divide each side by -5

x = 6/5

Now pick either equation and replace x with 6/5. I'll use equation 2:

3*(6/5) + y = 7

Multiply:

18/5 + y = 7

Subtract to isolate y:

y = 7 - 18/5

Common denominator:

y = 35/5 - 18/5

y = 17/5

WAY 2:

Substitute y in equation 2 for the answer from equation 1:

3x + 1 + 2x = 7

Combine like terms:

5x = 6

Divide by 5 to isolate x:

x = 6/5

Same as before - pick either equation and replace x with 6/5. I'll use equation 2:

3*(6/5) + y = 7

Multiply:

18/5 + y = 7

Subtract to isolate y:

y = 7 - 18/5

Common denominator:

y = 35/5 - 18/5

y = 17/5

Answer 29

y = 2x + 1

3x + y = 7

1 - Given these equations, we must equate one them to each other from the perspective of one of the variables. That will enable us to solve for the other. Given that the first equation is already y in terms of x and constant, let’s solve for y with the second equation:

1. 3x + y = 7
2. Subtract 3x from both sides.
3. y = -3x + 7

2 - Now we equate these equations to each other and solve for x.

1. 2x + 1 = -3x + 7
2. Add 3x to both sides.
3. 5x + 1 = 7
4. Subtract 1 from both sides.
5. 5x = 6
6. Divide both sides by 5
7. x = 6/5 or 1.2

3 - Now we take this x value of 6/5 and plug it back in to either one of the original equations to solve for y, and plug both values into the remaining equation to reverse check:

1. y = 2x + 1
2. Substitute x for 6/5
3. y = 2(6/5) + 1
4. 2 x 6/5 = 12/5 or 2.4
5. y = 12/5 + 1
6. 1 = 5/5
7. 12/5 + 5/5 = 17/5 or 3.4
8. y = 17/5 or 3.4
9. 3x + y = 7
10. Substitute x for 6/5 and y for 17/5
11. 3 (6/5) + 17/5 = 7
12. 3 x 6/5 = 18/5 or 3.6
13. 18/5 + 17/5 = 35/5
14. 3.6 + 3.4 = 7
15. 35/5 = 7
16. 7 = 7

Final Answer: [x = 6/5, y = 17/5] or [x = 1.2, y = 3.4]

Answer 30

Let’s solve step by step carefully:

We are given:

(1) y=2x+1y = 2x + 1y=2x+1

(2) 3x+y=73x + y = 73x+y=7

Step 1: Choose the method

Since equation (1) already expresses y in terms of x, the substitution method is the most efficient.

Step 2: Substitute equation (1) into equation (2)

Replace yyy in (2) with 2x+12x + 12x+1:

3x+(2x+1)=73x + (2x + 1) = 73x+(2x+1)=7 Step 3: Simplify

3x+2x+1=73x + 2x + 1 = 73x+2x+1=7 5x+1=75x + 1 = 75x+1=7 Step 4: Solve for x

5x=7−15x = 7 - 15x=7−1 5x=65x = 65x=6 x=65=1.2x = \frac{6}{5} = 1.2x=56​=1.2 Step 5: Substitute x back into equation (1) to find y

y=2(1.2)+1y = 2(1.2) + 1y=2(1.2)+1 y=2.4+1=3.4y = 2.4 + 1 = 3.4y=2.4+1=3.4

Final Answer:

x=1.2,y=3.4x = 1.2, \quad y = 3.4x=1.2,y=3.4 Method used: Substitution method.

Answer 31

Answer 32

Since equation 1 is already expressed as y = 2x+1, we can use the substitution method to solve this

Replace y in equation 2 with 2x+1

3x+(2x+1) = 7

Solve for x

5x+1 = 7

5x+1-1 = 7-1

5x = 6

x = 6/5

Now solve for y

y = 2(6/5)+1

y. = 12/5+5/5

y = 17/5

x = 6/5, y = 17/5

Answer 33

For this system, I would choose the substitution method because equation (1) already has y isolated (y = 2x + 1), which makes substitution very straightforward.

Solution:

Step 1: Substitute equation (1) into equation (2)

Since y = 2x + 1, I'll replace y in equation (2):

3x + y = 7 3x + (2x + 1) = 7

Step 2: Solve for x

Combine like terms:

1. 3x + 2x + 1 = 7
2. 5x + 1 = 7

Subtract 1 from both sides:

1. 5x = 6

Divide by 5:

1. x = 6/5 or x = 1.2

Step 3: Solve for y

Substitute x = 6/5 back into equation (1):

1. y = 2x + 1
2. y = 2(6/5) + 1
3. y = 12/5 + 5/5
4. y = 17/5 or y = 3.4

Step 4: Check the solution

Check in equation (2): 3x + y = 7

1. 3(6/5) + 17/5 = 18/5 + 17/5 = 35/5 = 7 ✓

Answer: x = 6/5, y = 17/5 (or x = 1.2, y = 3.4)

Why substitution? When one equation already has a variable isolated (like y = 2x + 1), substitution is usually the quickest and cleanest method. The elimination method would also work but would require an extra step of manipulating the equations first.

Answer 34

y = 2x + 1

3x + y = 7

Since one of the equations is already solved for y, it will be easiest to solve this system using Substitution. Replace the y in the second equation with the 2x + 1 from the first equation.

3x + (2x + 1) = 7

Now we have an equation with just one variable that we can solve. We don't need the parentheses since they don't have a coefficient.

3x + 2x + 1 = 7

Combine like terms on the left side of the equal sign.

5x + 1 = 7

Now we need to get the x by itself. Start by subtracting the 1 over the right side of the equation.

5x = 6

Divide both sides by 5.

x = 6/5

Now we can solve for y by plugging x= 6/5 back into either of the two original equations. Since y = 2x+1 is already solved for y, that one will be the easiest to work with.

y = 2(6/5) + 1

Multiply. We can turn any whole number into a fraction by putting it over 1. So we can 2 as 2/1 and then multiply straight across on the top and bottom of the fractions.

y = 12/5 + 1

In order to add fractions, we need a common denominator. We can write 1 as 1/1 and then multiply the top and bottom of the fraction by 5 to get a common denominator.

y = 12/5 +5/5

When we add fractions, we add the numerators but keep the denominator the same.

y=17/5

Our solution is

x = 6/5

y= 17/5

This can also be written as a coordinate pair: (6/5 , 17/5)

Answer 35

Step 1: Since y = 2x + 1 from equation (1), substitute this expression for y into equation (2):

3x + y = 7 3x + (2x + 1) = 7

Step 2: Simplify and solve for x:

3x + 2x + 1 = 7 5x + 1 = 7 5x = 6 x = 6/5

Step 3: Substitute x = 6/5 back into equation (1) to find y:

y = 2(6/5) + 1 y = 12/5 + 5/5 y = 17/5

Solution: x = 6/5 and y = 17/5

Answer 36

For this problem, one should use the substitution method. While other methods could be used, the substitution method is most convenient as one equation is already in the 'y=' form.

So, substituting equation 1 into equation 2, we have:

3x+2x+1=7

Combining like terms, we simplify the system to:

5x+1=7

Now, we have one equation with only one unknown variable, x, so we can solve for x as follows:

Subtract 1 from both sides to yield:

5x=6

Divide both sides by 5 to get x by itself:

x=6/5

Now that we have a value for x, we can plug x into either equation 1 or 2 to find a value for y. Solving a system of equations means that we are finding the coordinate where these two lines will cross. In other words, plugging x into either equation will result in the same y-value.

So, we will choose to plug x into equation 1, as it is already in the 'y=' form, and we are looking for the value of y.

Plugging x=6/5 into equation 1 results in:

y=2(6/5)+1

Then, multiplying the 2 by 6/5:

y=12/5+1

And finally, adding those terms together, while knowing that 1 is equal to 5/5:

y= 12/5 + 5/5

y=17/5

The final solution is: x=6/5 and y=17/5

If needed as a coordinate, the answer is (6/5, 17/5).

Answer 37

Solving system of two linear equations

Which method would you choose to solve the following system of equations? Solve it.

y=2x+1

3x+y=7 ​

You can choose from graphing, substitution, and elimination.

Since one of my equations is already in slope intercept form, I would choose substitution.

Substitute 2x + 1 in for y in the second equation

3x + (2x + 1) = 7

3x + 2x + 1 = 7 Distribute 1 across the parentheses

5x + 1 = 7

-1 -1 Subtract 1 from both sides

5x = 6 Divide both sides by 5

5 5

x = 6/5

Solve for y:

y = 2(6/5) +1

y = 12/5 + 5/5

y = 17/5

The solution set is (6/5,17/5)

You can check the solutions by plugging x and y into each equation to see if they are equal on both sides.

y = 2x + 1

17/5 = 2(6/5) + 1

17/5 = 12/5 +1

17/5 = 12/5 + 5/5

17/5 = 17/5

3x + y = 7

3(6/5) + 17/5 = 7

18/5 + 17/5 = 7

35/5 = 7

7 = 7

Answer 38

(1) y = 2x + 1

(2) 3x + y = 2

To solve this, we can apply the substitution method which would be easier in this case since the equations are in to different forms (slope-intercept form and standard form). If they were in the same form, then the elimination method would be easier to use. In the substitution method, you can substitute one equation into the other. So, lets substitute equation 1 into equation 2 since equation 1 has already been solved for y so we can substitute the entire equation into the y of the second equation like this:

3x + (2x + 1) = 2

What we just did was replace the y of the second equation with the first equation. Now we can simplify this by adding like terms (3x and 2x):

5x + 1 = 2

We can subtract 1 from both sides since we already added up like terms on the same side:

5x = 1

To solve for x, we can divide by 5 on both sides

x = 1/5

Now that we solved for x, we can solve for y by substituting this value of x into any equation (1 or 2). Lets plug it into the first equation:

y = 2(1/5) + 1

= 2/5 + 1

= 2/5 + 5/5

y = 7/5

So, our final answer as a coordinate point is (1/5, 7/5).

Answer 39

When one of the equations is already solved for a variable (like y=2x+1), the Substitution Method is usually the fastest and easiest choice.

We simply substitute (1) into (2):

3x+(2x+1)=7

Combine like terms:

3x+2x+1=7 ⇒ 5x+1=7

Solve for x:

5x=6 ⇒ x=6/5​

Plug back into (1) to find y:

y=2(6/5)+1 ⇒ 12/5+1 ⇒ 17/5

Final answer

x = 6/5 , y = 17/5

To check the result, just substitute x=6/5​ and y=17/5 into both original equations and verify that both sides are equal.

Answer 40

1. y = 2x + 1
2. 3x + y = 7

In these two equations there are two unknowns: x & y. Normally, algebra is all about an equation with one unknown - x - and you rearrange the equation to solve for x, but that's impossible in an equation with more than one unknown. Unless you change the equation so there's only one unknown.

With 1 & 2, x & y mean the same thing in both equations, and we can use that to make this solvable. See how equation 1 tells us that the unknown y is the same thing as the equation (2x + 1)? Well if y is the same in both equations, we can take y in equation 2 and replace it with (2x + 1) we get:

3x + (2x + 1) = 7

Now equation 2 only has one unknown with x. Since there's no multiplication with y the parentheses can just go away, and we can solve for x like a normal algebra equation.

3x + 2x + 1 = 7 -> 5x + 1 = 7 -> 5x = 6 -> x = 6/5 = 1.2

Now that we know what x is, we can go to either of the original equations, put our value for x in, and solve for y. Let's use equation 1, since it all just equals y anyways.

y = 2x + 1 -> y = 2(2.1) + 1 -> y = 4.2 + 1 -> y = 5.2

This method is called the Substitution Method, since you substitute one unknown with another to make it solvable.

Answer 41

To solve the given system of two linear equations, I would choose the substitution method.

The equations are:

1. y=2x+1
2. 3x+y=7

I chose the substitution method because the first equation is already solved for y, making it very simple to substitute the expression for y into the second equation.

Here is the step-by-step solution:

Step 1: Substitute the expression for y from equation (1) into equation (2).

The expression for y from equation (1) is 2x+1.

Substitute this into equation (2):

3x+(2x+1)=7

Step 2: Solve the new equation for x.

3x+2x+1=7

5x+1=7

5x=7−1

5x=6

x=6/5

Step 3: Substitute the value of x back into equation (1) to solve for y.

Now that we have the value for x, we can substitute it back into the first equation to find y.

y=2x+1

y=2(6/5​)+(5/5)

y=(12/5)+(5/5)

To add these numbers, find a common denominator:

y=(12/5) ​+ (5/5)​

y=17/5

Answer (6/5, 17/5)

Answer 42

Many responses have clarified the best method to use, so I will simply add a slightly more generic way to think about such problems -

Say you have a system of equations represented by -

Ax + By = C (1)

Dx + Ey = F (2)

To use the substitution method most effectively, simply bring the coefficients of any one variable to be identical so that you can subtract the two equations resulting in an equation with only one variable. The reduction in number of variables is the key concept you need to visualize and learn.

In this specific case, you can multiply equation (1) by D and equation 2 by A like so

(D*A)x + (D*B)y = (D*C) (1a)

(A*D)x + (A*E)y = (A*F) (2a)

Now subtract (2a) from (1a) which will cancel out the X term (using the commutative property of multiplication obviously), leaving you with

(DB - AE) y = DC - AF

or y = (DC - AF) / (DB - AE) (3)

In our specific example, which I will rewrite below

y=2x+1 (1)

3x+y=7 ​ (2)

We can rewrite each equation as below and derive the values that map to A, B, C and D, E, F in the generalized solution above -

y=2x+1 (1) => -2x + y = 1 => A = -2, B = 1, C = 1

3x+y=7 ​ (2) => 3x + y = 7 => D = 3, E = 1, F = 7

Plugging in these values into the solution per equation (3) above -

y = (DC - AF) / (DB - AE)

or y = (3*1 - (-2)*7) / (3*1 - (-2)*1)

or y = (3 + 14) / (3 + 2)

or y = 17/5

Substituting the value of y in equation (1) will give us the value of X per below

y=2x+1 (1)

or x = (y - 1)/2

or x = (17/5 - 1) / 2 = 6/5

This gives us the answer x = 6/5 and y = 17/5

And the generic method can even give you an easy way to write a program that can solve such systems of equations!

Answer 43

y = 2x + 1

3x + y = 7

Using the substitution method is the best way to solve the problem.

Note that the first equation is already written in terms of y.

Place 2x + 1 in place of the y in the second equation:

3x + 2x + 1 = 7

Step 1: First combine the like terms: 5x + 1 = 7

Step 2: subtract 1 from both sides of the equation (this will isolate the term with the variable):

5x +1 -1 = 7 - 1

5x = 6

Step 3: To isolate x, divide both sides by 5:

5/5 x = 6/5 (5/5 = 1)

x = 6/5

Step 4: Put x = 6/5 in place of x in the first equation to solve for y:

y = 2(6/5) + 1

y = 12/5 + 1

y = 12/5 + 5/5

y = 17/5

As an ordered pair, the answer is (6/5, 17/5)

Answer 44

y = 2x + 1

3x + y = 7

Using the substitution method would be the most straightforward way to solve this system since one of the equations is already in terms of a variable.

In this case, the first equation is already written in terms of y.

We can substitute 2x + 1 in place of the y in the second equation:

3x + 2x + 1 = 7

To solve:

First combine like terms:

5x + 1 = 7

Next, subtract 1 from both sides of the equation (this will isolate the term with the variable):

5x +1 -1 = 7 - 1

5x = 6

To isolate x, divide both sides by 5:

5/5 x = 6/5

x = 6/5

Now substitute 6/5 in place of x in the first equation to solve for y:

y = 2(6/5) + 1

y = 12/5 + 1

y = 12/5 + 5/5

y = 17/5

As an ordered pair, the answer is (6/5, 17/5)

You can always check your work by substituting the answer back into both equations to make sure they are both true:

y = 2x + 1

17/5 = 2(6/5) + 1

17/5 = 12/5 +1

17/5 = 12/5 + 5/5

17/5 = 17/5

3x + y = 7

3(6/5) + 17/5 = 7

18/5 + 17/5 = 7

35/5 = 7

7 = 7