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

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 6

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 7

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 8

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 9

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 10

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 11

(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 12

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 13

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 14

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 15

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 16

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 17

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