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# Systems of Equations—Solving with Substitution Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Systems of Equations—Solving with Substitution Study Guide.

## Systems of Equations—Solving with Substitution Practice Questions

### Problems

#### Practice 1

Solve each system of equations.

1. x – 2y = –10, 6x + 2y = –4
2. 3y – 5x = 6, yx = 4
3. 8x – 4 = 2y, –5x + 3y = –13

#### Practice 2

Find the solution to this system of equations:

3z + 2x = y – 1
5y – 2z = x + 2
3x = –4z

### Solutions

#### Practice 1

 1. Start by writing one variable in terms of the other. Because x has no coefficient in the first equation, start by writing x in terms of y using that equation: x – 2y = –10 x = 2y – 10 Replace x in the second equation with the expression that is equal to x, 2y – 10: 6(2y – 10) + 2y = –4 Solve for the value of y: 6(2y – 10) + 2y = –4 12y – 60 + 2y = –4 14y – 60 = –4 14y = 56 y = 4 Replace y with its value in either equation and solve for the value of x: x – 2(4) = –10 x – 8 = –10 x = –2 The solution to this system of equations is x = –2, y = 4. 2. Start by writing one variable in terms of the other. Because y has no coefficient in the second equation, start by writing y in terms of x using that equation: y – x = 4 y = x + 4 Replace y in the first equation with the expression that is equal to y, x + 4: 3(x + 4) – 5x = 6 Solve for the value of x 3(x + 4) – 5x = 6 3x + 12 – 5x = 6 –2x + 12 = 6 –2x = –6 x = 3 Replace x with its value in either equation and solve for the value of y: y – x = 4 y – 3 = 4 y = 7 The solution to this system of equations is x = 3, y = 7. 3. Start by writing one variable in terms of the other. Write y in terms of x using the first equation. Divide both sides of the equation by 2: 8x – 4 = 2y 4x – 2 = y Replace y in the second equation with the expression that is equal to y, 4x – 2: –5x + 3(4x – 2) = –13 Solve for the value of x: –5x + 3(4x – 2) = –13 –5x + 12x – 6 = –13 7x – 6 = –13 7x = –7 x = –1 Replace x with its value in either equation and solve for the value of y: 8(–1) – 4 = 2y –8 – 4 = 2y –12 = 2y –6 = y The solution to this system of equations is x = –1, y = –6.

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