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# Systems of Equations—Solving with Substitution Practice Questions (page 2)

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Updated on Oct 3, 2011

#### Practice 2

Start by writing one variable in terms of two other variables. Write y in terms of x and z:

3z + 2x = y – 1
3z + 2x + 1 = y

Next, replace y in the second equation with the expression 3z + 2x + 1. The third equation does not have a y at all.

5y – 2z = x + 2
5(3z + 2x + 1) – 2z = x + 2
15z + 10x + 5 – 2z = x + 2
13z + 10x + 5 =x+ 2
13z + 9x + 5 = 2
13z + 9x = –3

There are now two equations and two variables:

13z + 9x = –3
3x = –4z

Write one variable in terms of the other. Write x in terms of z using the second equation:

3x = –4z

Replace x in the first equation with the expression that is equal to x,

13z + 9x = –3
13z – 12z = –3
z = –3

Replace z with its value in either equation and solve for the value of x:

3x = –4z
3x = –4(–3)
3x = 12
x = 4

Substitute the values of x and z into any one of the original equations to find the value of y:

3z + 2x = y – 1
3(–3) + 2(4) = y – 1
–9 + 8 = y – 1
–1 = y – 1
y = 0

The solution to this system of equations is x = 4, y = 0, z = –3.

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