Cramer's Rule practice worksheet with 12 systems of two equations to solve.

Here are the solutions for each system of equations using Cramer's Rule.

How to use Cramer's Rule:
For a system like:
$ax + by = e$
$cx + dy = f$

1. Find the main determinant ($D$): $(a \cdot d) - (b \cdot c)$
2. Find $D_x$ (replace x-column with answers): $(e \cdot d) - (b \cdot f)$
3. Find $D_y$ (replace y-column with answers): $(a \cdot f) - (e \cdot c)$
4. Solve: $x = D_x / D$ and $y = D_y / D$

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1)
$D = (-2)(4) - (-2)(4) = -8 - (-8) = 0$
Since $D = 0$, there is no unique solution.

2)
$D = (-5)(-3) - (-2)(3) = 15 - (-6) = 21$
$D_x = (-7)(-3) - (-2)(-21) = 21 - 42 = -21$
$D_y = (-5)(-21) - (-7)(3) = 105 - (-21) = 126$
$x = -21 / 21 = -1$
$y = 126 / 21 = 6$

3)
$D = (4)(3) - (4)(3) = 12 - 12 = 0$
Since $D = 0$, there is no unique solution.

4)
$D = (4)(4) - (3)(3) = 16 - 9 = 7$
$D_x = (-26)(4) - (3)(-30) = -104 - (-90) = -14$
$D_y = (4)(-30) - (-26)(3) = -120 - (-78) = -42$
$x = -14 / 7 = -2$
$y = -42 / 7 = -6$

5)
$D = (-2)(5) - (6)(6) = -10 - 36 = -46$
$D_x = (4)(5) - (6)(-35) = 20 - (-210) = 230$
$D_y = (-2)(-35) - (4)(6) = 70 - 24 = 46$
$x = 230 / -46 = -5$
$y = 46 / -46 = -1$

6)
$D = (-3)(2) - (-2)(3) = -6 - (-6) = 0$
Since $D = 0$, there is no unique solution.

7)
$D = (-3)(2) - (-5)(4) = -6 - (-20) = 14$
$D_x = (-1)(2) - (-5)(20) = -2 - (-100) = 98$
$D_y = (-3)(20) - (-1)(4) = -60 - (-4) = -56$
$x = 98 / 14 = 7$
$y = -56 / 14 = -4$

8)
$D = (6)(1) - (6)(1) = 6 - 6 = 0$
Since $D = 0$, there is no unique solution.

9)
$D = (-1)(-3) - (-2)(-2) = 3 - 4 = -1$
$D_x = (-9)(-3) - (-2)(-11) = 27 - 22 = 5$
$D_y = (-1)(-11) - (-9)(-2) = 11 - 18 = -7$
$x = 5 / -1 = -5$
$y = -7 / -1 = 7$

10)
$D = (-6)(-5) - (-3)(-1) = 30 - 3 = 27$
$D_x = (-15)(-5) - (-3)(-16) = 75 - 48 = 27$
$D_y = (-6)(-16) - (-15)(-1) = 96 - 15 = 81$
$x = 27 / 27 = 1$
$y = 81 / 27 = 3$

11)
$D = (-5)(-6) - (6)(2) = 30 - 12 = 18$
$D_x = (60)(-6) - (6)(-42) = -360 - (-252) = -108$
$D_y = (-5)(-42) - (60)(2) = 210 - 120 = 90$
$x = -108 / 18 = -6$
$y = 90 / 18 = 5$

12)
$D = (-4)(-2) - (-4)(-2) = 8 - 8 = 0$
Since $D = 0$, there is no unique solution.

Final Answer:
1) No unique solution
2) x = -1, y = 6
3) No unique solution
4) x = -2, y = -6
5) x = -5, y = -1
6) No unique solution
7) x = 7, y = -4
8) No unique solution
9) x = -5, y = 7
10) x = 1, y = 3
11) x = -6, y = 5
12) No unique solution