Grade 8 Algebra: Solving Systems by Substitution

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Level 1

One variable already isolated

1. $(4, 8)$
2. $(3, 6)$
3. $(12, 4)$
4. $(3, 12)$
5. $(4, 2)$
6. $(7, 2)$
7. $(4, 5)$
8. $(3, 7)$

Name: ______________________Date: ____________Score: ____ / 8

Negatives and coefficients

1. Solve by substitution:
$\begin{cases} y = -2x + 3 \\ 3x + y = 5 \end{cases}$
2. Solve by substitution:
$\begin{cases} y = x - 4 \\ 2x + 3y = 3 \end{cases}$
3. Solve by substitution:
$\begin{cases} x = 2y - 1 \\ 3x - y = 7 \end{cases}$
4. Solve by substitution:
$\begin{cases} y = 3x + 2 \\ 4x - y = -5 \end{cases}$
5. Solve by substitution:
$\begin{cases} x = -y + 6 \\ 2x - 3y = -3 \end{cases}$
6. Solve by substitution:
$\begin{cases} y = -x - 1 \\ 5x + 2y = 4 \end{cases}$
7. Solve by substitution:
$\begin{cases} y = 2x - 3 \\ y = -x + 6 \end{cases}$
8. Solve by substitution:
$\begin{cases} x = 4y \\ 2x - 5y = -6 \end{cases}$

1. $(2, -1)$
2. $(3, -1)$
3. $(3, 2)$
4. $(-3, -7)$
5. $(3, 3)$
6. $(2, -3)$
7. $(3, 3)$
8. $(-8, -2)$

Name: ______________________Date: ____________Score: ____ / 8

Isolate first; special cases

1. Solve by substitution:
$\begin{cases} x + y = 7 \\ 2x - y = 2 \end{cases}$
2. Solve by substitution:
$\begin{cases} x - 2y = 1 \\ 3x + y = 10 \end{cases}$
3. Solve by substitution:
$\begin{cases} 2x + y = -1 \\ 4x - 3y = 13 \end{cases}$
4. Solve by substitution:
$\begin{cases} y = 2x + 5 \\ 4x - 2y = 6 \end{cases}$
5. Solve by substitution:
$\begin{cases} y = 3x - 1 \\ 6x - 2y = 2 \end{cases}$
6. Solve by substitution:
$\begin{cases} x + 3y = 4 \\ 2x - y = -6 \end{cases}$
7. Solve by substitution:
$\begin{cases} 3x + 2y = 8 \\ x - y = 6 \end{cases}$
8. Solve by substitution:
$\begin{cases} y = \tfrac{1}{2}x + 2 \\ x + 2y = 12 \end{cases}$