Solve for 𝑥 and 𝑦 in the system of equations: 2 𝑥 + 3 𝑦 − 4 = 5 𝑥 − 2 𝑦 + 7 and 4 ( 𝑥 − 1 ) + 2 ( 𝑦 + 3 ) = 3 𝑥 − 𝑦 + 10

Question

This algebra problem requires solving a system of two linear equations with two variables, x and y. The equations are given as:

2x + 3y − 4 = 5x − 2y + 7

4(x−1) + 2(y+3) = 3x − y + 10

Answer 1

Problem

Given the system of equations ...

1) 2x + 3y - 4 = 5x - 2y + 7

2) 4(x - 1) + 2(y + 3) = 3x - y + 10

... follow the steps below to solve for x and y:

Method of Substitution

Step 1

Rearrange and combine like terms in equation 1 into the form Ax + By = C:

2x + 3y - 4 = 5x - 2y + 7

2x - 5x + 3y + 2y = 7 + 4

-3x + 5y = 11

Step 2

Use the distributive property and rearrange terms to simplify equation 2:

4(x - 1) + 2(y + 3) = 3x - y + 10

4x - 4 + 2y + 6 = 3x - y + 10

4x - 3x + 2y + y = 10 + 4 - 6

x + 3y = 8

Step 3

Isolate x or y on one side of equation 2 (here we are using x) to get the following substitution:

x = 8 - 3y

Step 4

Plug in the substitution we solved for in step 3 into each x in equation 1:

-3x + 5y = 11 becomes -3(8 - 3y) + 5y = 11

-24 + 9y + 5y = 11

-24 + 14y = 11

14y = 35

y = 5/2 or 2.5

Step 5

Solve for x by plugging in the y we found in step 4 into equation 2:

x + 3y = 8 becomes x + 3(2.5) = 8

x + 7.5 = 8

x = 1/2 or 0.5

Step 6

Check work to make sure answers line up with our equations:

-3x + 5y becomes -3(0.5) + 5(2.5)

-3(0.5) + 5(2.5) = 11 ✔

x + 3y becomes 0.5 + 3(2.5)

0.5 + 3(2.5) = 8 ✔

Solution

x = 1/2 or 0.5, y = 5/2 or 2.5

Answer 2

2x + 3y − 4 = 5x − 2y + 7 (1)

4(x−1) + 2(y+3) = 3x − y + 10 (2)

Step 1: Re-arrange equations (1) and (2) into standard form, Ax + By = C, where we can apply the process of elimination. By adding adding a quantity to both sides of the equation, we can write the equations in standard form:

Equation (1):

2x - 5x + 3y + 2y − 4 + 4 = 5x - 5x − 2y + 2y + 7 + 4

-3x + 5y = 11 (1)

Equation (2):

4x - 4 + 2y + 6 = 3x - y + 10

4x - 3x + 2y + y + 2 - 2 = 3x - 3x - y + y + 10 - 2

x + 3y = 8 (2)

We now have our system of equations

-3x + 5y = 11 (1)

x + 3y = 8 (2)

Step 2: With our system of equations from step 1, we shall eliminate an unknown variable (either x or y) from the system of equations. We cannot eliminate either variable by adding the equations as they are, so we must multiply one of the equations by a coefficient that will allow them to cancel either other out. We can see that multiplying equation (2) by 3 will eliminate the x coefficient, allowing us to solve for y:

-3x + 5y = 11

3*(x + 3y) = 3*(8)

-3x + 5y = 11 (1)

3x + 9y = 24 (2)

The coefficients of the x terms are now opposites. We can now add both equations to eliminate x:

-3x + 5y = 11

3x + 9y = 24

14y = 35

Step 3: We shall now solve for y, by re-arranging the equation from step 2.

14y = 35

y = 35/14

y = 5/2 or 2.5

Step 4: We can now solve for x, by substituting our known value of y = 2.5 into either equation (1) or (2). Let's apply the substitution to both equations, in order to check our work.

Equation (1):

-3x + 5y = 11

-3x + 5*(2.5) = 11

-3x + 12.5 = 11

-3x + 12.5 - 12.5 = 11 - 12.5

-3x = -1.5

x = 0.5

Equation (2):

x + 3y = 8

x + 3*(2.5) = 8

x + 7.5 = 8

x + 7.5 - 7.5 = 8 - 7.5

x = 0.5

Answer:

x = 0.5, y = 2.5

*Note: It is important to choose coefficients in step 2 to multiply that will result in easier math, to reduce the risk of errors or spending excess time on a test. Multiplying equation (2) by 3 is the obvious easier path to a solution, but this may not always be the case. Depending on the coefficients assigned to the variables in your problem, you may have to multiply both equations (1) and (2) by coefficients.

Answer 3

This problem assesses students' ability to simplify, combine like terms, and solve systems of linear equations using methods like substitution or elimination. It provides a deeper challenge and reinforces advanced algebraic skills.

Steps to Solve:

1. Simplify each equation by distributing and combining like terms:

For the first equation:

2x + 3y − 4 = 5x − 2y + 7

2x + 3y − 4 − 5x + 2y = 7

For the second equation:

4(x−1) + 2(y+3) = 3x − y + 10

4x − 4 + 2y + 6 = 3x − y + 10

4x + 2y + 2 = 3x − y + 10

x + 3y = 8

1. Solve the system of equations using substitution or elimination:

System of equations:

−3x + 5y = 11

x + 3y = 8

Using the elimination method, you can multiply the second equation by 3 to align the x terms for elimination:

3(x+3y) = 3x8

3x + 9y = 24

Add this to the first equation to eliminate x:

−3x + 5y = 11

+3x + 9y = 24

14y = 35

y = 35/14

y = 2.5

Substitute y back into the second simplified equation to find x:

x + 3(2.5) = 8

x + 7.5 = 8

x = 0.5

So, the solution to the system of equations is x = 0.5 and y = 2.5.