Worksheet for solving systems of equations using graphing, substitution, and elimination methods.
Graphing Systems of Equations Worksheet featuring two problems with linear equations to be solved by graphing, substitution, and elimination methods.
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Step-by-step solution for: Graphing Systems of Equations Worksheets - Math Monks
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Show Answer Key & Explanations
Step-by-step solution for: Graphing Systems of Equations Worksheets - Math Monks
Let's solve each system of equations step by step, as per the worksheet instructions.
---
We'll analyze and solve each system graphically. Since we can't draw on this platform, I’ll describe how to graph each line and find their intersection point (solution).
---
#### Problem 1:
$$
y = -\frac{1}{2}x - 2 \\
y = -\frac{1}{2}x + 2
$$
Analysis:
- Both lines have the same slope: $-\frac{1}{2}$.
- But different y-intercepts: -2 and +2.
- So, these lines are parallel and will never intersect.
✔ Solution: No solution (inconsistent system)
> ✔ Graph: Draw two parallel lines with slope $-\frac{1}{2}$, one crossing y-axis at $-2$, the other at $+2$. They never meet.
---
#### Problem 2:
$$
y = -\frac{1}{4}x - 2 \\
y = x + 3
$$
Step-by-step graphing:
- First equation: $y = -\frac{1}{4}x - 2$
- Slope: $-\frac{1}{4}$, y-intercept: $-2$
- Plot point $(0, -2)$, then go down 1, right 4 → next point $(4, -3)$
- Second equation: $y = x + 3$
- Slope: $1$, y-intercept: $3$
- Plot $(0, 3)$, then up 1, right 1 → $(1, 4)$
Now find intersection point algebraically for accuracy:
Set equations equal:
$$
-\frac{1}{4}x - 2 = x + 3
$$
Multiply both sides by 4 to eliminate fraction:
$$
- x - 8 = 4x + 12
$$
Bring all terms to one side:
$$
- x - 8 - 4x - 12 = 0 \Rightarrow -5x - 20 = 0
\Rightarrow -5x = 20 \Rightarrow x = -4
$$
Plug into $y = x + 3$:
$y = -4 + 3 = -1$
✔ Solution: $(-4, -1)$
> ✔ Graph: Lines cross at $(-4, -1)$
---
---
#### Problem 3:
$$
x = -2y \quad \text{(1)} \\
x - y = 9 \quad \text{(2)}
$$
Substitute (1) into (2):
$$
(-2y) - y = 9 \Rightarrow -3y = 9 \Rightarrow y = -3
$$
Now plug back into (1):
$x = -2(-3) = 6$
✔ Solution: $(6, -3)$
---
#### Problem 4:
$$
3x - 4y = 10 \quad \text{(1)} \\
2y + 4x = 6 \quad \text{(2)}
$$
Solve (2) for $y$:
$$
2y = -4x + 6 \Rightarrow y = -2x + 3
$$
Substitute into (1):
$$
3x - 4(-2x + 3) = 10 \\
3x + 8x - 12 = 10 \Rightarrow 11x = 22 \Rightarrow x = 2
$$
Now $y = -2(2) + 3 = -4 + 3 = -1$
✔ Solution: $(2, -1)$
---
#### Problem 5:
$$
y = x + 5 \quad \text{(1)} \\
y = -2x - 4 \quad \text{(2)}
$$
Set equal:
$$
x + 5 = -2x - 4 \\
x + 2x = -4 - 5 \Rightarrow 3x = -9 \Rightarrow x = -3
$$
Then $y = -3 + 5 = 2$
✔ Solution: $(-3, 2)$
---
---
#### Problem 6:
$$
8x + y = -16 \quad \text{(1)} \\
-3x + y = -5 \quad \text{(2)}
$$
Subtract (2) from (1) to eliminate $y$:
$$
(8x + y) - (-3x + y) = -16 - (-5) \\
8x + y + 3x - y = -16 + 5 \Rightarrow 11x = -11 \Rightarrow x = -1
$$
Plug into (2):
$-3(-1) + y = -5 \Rightarrow 3 + y = -5 \Rightarrow y = -8$
✔ Solution: $(-1, -8)$
---
#### Problem 7:
$$
5x + 4y = -14 \quad \text{(1)} \\
3x + 6y = 6 \quad \text{(2)}
$$
Make coefficients of $x$ or $y$ match. Let's eliminate $x$.
Multiply (1) by 3: $15x + 12y = -42$
Multiply (2) by 5: $15x + 30y = 30$
Now subtract:
$$
(15x + 30y) - (15x + 12y) = 30 - (-42) \\
15x + 30y - 15x - 12y = 72 \Rightarrow 18y = 72 \Rightarrow y = 4
$$
Plug into (1):
$5x + 4(4) = -14 \Rightarrow 5x + 16 = -14 \Rightarrow 5x = -30 \Rightarrow x = -6$
✔ Solution: $(-6, 4)$
---
#### Problem 8:
$$
-6x + 5y = -1 \quad \text{(1)} \\
6x + 4y = -10 \quad \text{(2)}
$$
Add equations directly (eliminate $x$):
$$
(-6x + 5y) + (6x + 4y) = -1 + (-10) \\
0x + 9y = -11 \Rightarrow y = -\frac{11}{9}
$$
Now plug into (1):
$-6x + 5(-\frac{11}{9}) = -1 \Rightarrow -6x - \frac{55}{9} = -1$
Multiply everything by 9:
$$
-54x - 55 = -9 \Rightarrow -54x = 46 \Rightarrow x = -\frac{46}{54} = -\frac{23}{27}
$$
✔ Solution: $\left(-\frac{23}{27}, -\frac{11}{9}\right)$
---
| Problem | Method | Solution |
|--------|----------------|------------------------------|
| 1 | Graphing | No solution (parallel lines) |
| 2 | Graphing | $(-4, -1)$ |
| 3 | Substitution | $(6, -3)$ |
| 4 | Substitution | $(2, -1)$ |
| 5 | Substitution | $(-3, 2)$ |
| 6 | Elimination | $(-1, -8)$ |
| 7 | Elimination | $(-6, 4)$ |
| 8 | Elimination | $\left(-\frac{23}{27}, -\frac{11}{9}\right)$ |
---
Let me know if you'd like help graphing any of these or want visual descriptions!
---
Part 1: Solve each system by graphing
We'll analyze and solve each system graphically. Since we can't draw on this platform, I’ll describe how to graph each line and find their intersection point (solution).
---
#### Problem 1:
$$
y = -\frac{1}{2}x - 2 \\
y = -\frac{1}{2}x + 2
$$
Analysis:
- Both lines have the same slope: $-\frac{1}{2}$.
- But different y-intercepts: -2 and +2.
- So, these lines are parallel and will never intersect.
✔ Solution: No solution (inconsistent system)
> ✔ Graph: Draw two parallel lines with slope $-\frac{1}{2}$, one crossing y-axis at $-2$, the other at $+2$. They never meet.
---
#### Problem 2:
$$
y = -\frac{1}{4}x - 2 \\
y = x + 3
$$
Step-by-step graphing:
- First equation: $y = -\frac{1}{4}x - 2$
- Slope: $-\frac{1}{4}$, y-intercept: $-2$
- Plot point $(0, -2)$, then go down 1, right 4 → next point $(4, -3)$
- Second equation: $y = x + 3$
- Slope: $1$, y-intercept: $3$
- Plot $(0, 3)$, then up 1, right 1 → $(1, 4)$
Now find intersection point algebraically for accuracy:
Set equations equal:
$$
-\frac{1}{4}x - 2 = x + 3
$$
Multiply both sides by 4 to eliminate fraction:
$$
- x - 8 = 4x + 12
$$
Bring all terms to one side:
$$
- x - 8 - 4x - 12 = 0 \Rightarrow -5x - 20 = 0
\Rightarrow -5x = 20 \Rightarrow x = -4
$$
Plug into $y = x + 3$:
$y = -4 + 3 = -1$
✔ Solution: $(-4, -1)$
> ✔ Graph: Lines cross at $(-4, -1)$
---
Part 2: Solve using substitution method
---
#### Problem 3:
$$
x = -2y \quad \text{(1)} \\
x - y = 9 \quad \text{(2)}
$$
Substitute (1) into (2):
$$
(-2y) - y = 9 \Rightarrow -3y = 9 \Rightarrow y = -3
$$
Now plug back into (1):
$x = -2(-3) = 6$
✔ Solution: $(6, -3)$
---
#### Problem 4:
$$
3x - 4y = 10 \quad \text{(1)} \\
2y + 4x = 6 \quad \text{(2)}
$$
Solve (2) for $y$:
$$
2y = -4x + 6 \Rightarrow y = -2x + 3
$$
Substitute into (1):
$$
3x - 4(-2x + 3) = 10 \\
3x + 8x - 12 = 10 \Rightarrow 11x = 22 \Rightarrow x = 2
$$
Now $y = -2(2) + 3 = -4 + 3 = -1$
✔ Solution: $(2, -1)$
---
#### Problem 5:
$$
y = x + 5 \quad \text{(1)} \\
y = -2x - 4 \quad \text{(2)}
$$
Set equal:
$$
x + 5 = -2x - 4 \\
x + 2x = -4 - 5 \Rightarrow 3x = -9 \Rightarrow x = -3
$$
Then $y = -3 + 5 = 2$
✔ Solution: $(-3, 2)$
---
Part 3: Solve using elimination method
---
#### Problem 6:
$$
8x + y = -16 \quad \text{(1)} \\
-3x + y = -5 \quad \text{(2)}
$$
Subtract (2) from (1) to eliminate $y$:
$$
(8x + y) - (-3x + y) = -16 - (-5) \\
8x + y + 3x - y = -16 + 5 \Rightarrow 11x = -11 \Rightarrow x = -1
$$
Plug into (2):
$-3(-1) + y = -5 \Rightarrow 3 + y = -5 \Rightarrow y = -8$
✔ Solution: $(-1, -8)$
---
#### Problem 7:
$$
5x + 4y = -14 \quad \text{(1)} \\
3x + 6y = 6 \quad \text{(2)}
$$
Make coefficients of $x$ or $y$ match. Let's eliminate $x$.
Multiply (1) by 3: $15x + 12y = -42$
Multiply (2) by 5: $15x + 30y = 30$
Now subtract:
$$
(15x + 30y) - (15x + 12y) = 30 - (-42) \\
15x + 30y - 15x - 12y = 72 \Rightarrow 18y = 72 \Rightarrow y = 4
$$
Plug into (1):
$5x + 4(4) = -14 \Rightarrow 5x + 16 = -14 \Rightarrow 5x = -30 \Rightarrow x = -6$
✔ Solution: $(-6, 4)$
---
#### Problem 8:
$$
-6x + 5y = -1 \quad \text{(1)} \\
6x + 4y = -10 \quad \text{(2)}
$$
Add equations directly (eliminate $x$):
$$
(-6x + 5y) + (6x + 4y) = -1 + (-10) \\
0x + 9y = -11 \Rightarrow y = -\frac{11}{9}
$$
Now plug into (1):
$-6x + 5(-\frac{11}{9}) = -1 \Rightarrow -6x - \frac{55}{9} = -1$
Multiply everything by 9:
$$
-54x - 55 = -9 \Rightarrow -54x = 46 \Rightarrow x = -\frac{46}{54} = -\frac{23}{27}
$$
✔ Solution: $\left(-\frac{23}{27}, -\frac{11}{9}\right)$
---
✔ Final Answers Summary:
| Problem | Method | Solution |
|--------|----------------|------------------------------|
| 1 | Graphing | No solution (parallel lines) |
| 2 | Graphing | $(-4, -1)$ |
| 3 | Substitution | $(6, -3)$ |
| 4 | Substitution | $(2, -1)$ |
| 5 | Substitution | $(-3, 2)$ |
| 6 | Elimination | $(-1, -8)$ |
| 7 | Elimination | $(-6, 4)$ |
| 8 | Elimination | $\left(-\frac{23}{27}, -\frac{11}{9}\right)$ |
---
Let me know if you'd like help graphing any of these or want visual descriptions!
Parent Tip: Review the logic above to help your child master the concept of solving systems by graphing worksheet 6 1.