Unit 1 Worksheet 5 Solving Systems Of Equations In 2 Variables ... - Free Printable

Problem: Solving Systems of Equations in 2 Variables

The task involves solving systems of linear equations in two variables using various methods, including graphing, substitution, and the combination (elimination) method. Below, I will solve each system step by step and explain the solution process.

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#### Section 1: Solve Each of the Following Systems of Equations by Graphing

1. System:
\[
\begin{cases}
x - y = -1 \\
2x + y = 8
\end{cases}
\]

- Step 1: Rewrite each equation in slope-intercept form ($y = mx + b$).
- From $x - y = -1$:
\[
y = x + 1
\]
- From $2x + y = 8$:
\[
y = -2x + 8
\]

- Step 2: Graph both lines.
- The first line, $y = x + 1$, has a slope of 1 and a y-intercept of 1.
- The second line, $y = -2x + 8$, has a slope of -2 and a y-intercept of 8.

- Step 3: Find the point of intersection.
- By graphing, the lines intersect at the point $(3, 4)$.

- Solution:
\[
\boxed{(3, 4)}
\]

2. System:
\[
\begin{cases}
3x - 2y = 6 \\
5x + 3y = 21
\end{cases}
\]

- Step 1: Rewrite each equation in slope-intercept form.
- From $3x - 2y = 6$:
\[
y = \frac{3}{2}x - 3
\]
- From $5x + 3y = 21$:
\[
y = -\frac{5}{3}x + 7
\]

- Step 2: Graph both lines.
- The first line, $y = \frac{3}{2}x - 3$, has a slope of $\frac{3}{2}$ and a y-intercept of -3.
- The second line, $y = -\frac{5}{3}x + 7$, has a slope of $-\frac{5}{3}$ and a y-intercept of 7.

- Step 3: Find the point of intersection.
- By graphing, the lines intersect at the point $(3, 1.5)$.

- Solution:
\[
\boxed{(3, 1.5)}
\]

3. System:
\[
\begin{cases}
3x + y = -2 \\
6x + 2y = 4
\end{cases}
\]

- Step 1: Rewrite each equation in slope-intercept form.
- From $3x + y = -2$:
\[
y = -3x - 2
\]
- From $6x + 2y = 4$:
\[
y = -3x + 2
\]

- Step 2: Graph both lines.
- Both lines have the same slope ($-3$) but different y-intercepts, so they are parallel.

- Step 3: Determine if there is an intersection.
- Since the lines are parallel, they do not intersect.

- Solution:
\[
\boxed{\text{No solution}}
\]

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#### Section 2: Solve Each of the Following Systems of Equations Using the Substitution Method

1. System:
\[
\begin{cases}
x - y = 3 \\
x + y = 7
\end{cases}
\]

- Step 1: Solve one equation for one variable.
- From $x - y = 3$:
\[
x = y + 3
\]

- Step 2: Substitute into the other equation.
- Substitute $x = y + 3$ into $x + y = 7$:
\[
(y + 3) + y = 7
\]
\[
2y + 3 = 7
\]
\[
2y = 4
\]
\[
y = 2
\]

- Step 3: Solve for the other variable.
- Substitute $y = 2$ into $x = y + 3$:
\[
x = 2 + 3 = 5
\]

- Solution:
\[
\boxed{(5, 2)}
\]

2. System:
\[
\begin{cases}
3x + y = 8 \\
-x + 2y = 12
\end{cases}
\]

- Step 1: Solve one equation for one variable.
- From $3x + y = 8$:
\[
y = 8 - 3x
\]

- Step 2: Substitute into the other equation.
- Substitute $y = 8 - 3x$ into $-x + 2y = 12$:
\[
-x + 2(8 - 3x) = 12
\]
\[
-x + 16 - 6x = 12
\]
\[
-7x + 16 = 12
\]
\[
-7x = -4
\]
\[
x = \frac{4}{7}
\]

- Step 3: Solve for the other variable.
- Substitute $x = \frac{4}{7}$ into $y = 8 - 3x$:
\[
y = 8 - 3\left(\frac{4}{7}\right)
\]
\[
y = 8 - \frac{12}{7}
\]
\[
y = \frac{56}{7} - \frac{12}{7}
\]
\[
y = \frac{44}{7}
\]

- Solution:
\[
\boxed{\left(\frac{4}{7}, \frac{44}{7}\right)}
\]

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#### Section 3: Solve Each of the Following Systems of Equations Using the Combination (Elimination) Method

1. System:
\[
\begin{cases}
3x + 4y = 10 \\
5x + 6y = 14
\end{cases}
\]

- Step 1: Eliminate one variable by making coefficients equal.
- Multiply the first equation by 3 and the second equation by 2:
\[
\begin{cases}
9x + 12y = 30 \\
10x + 12y = 28
\end{cases}
\]

- Step 2: Subtract the equations to eliminate $y$.
\[
(10x + 12y) - (9x + 12y) = 28 - 30
\]
\[
x = -2
\]

- Step 3: Solve for the other variable.
- Substitute $x = -2$ into $3x + 4y = 10$:
\[
3(-2) + 4y = 10
\]
\[
-6 + 4y = 10
\]
\[
4y = 16
\]
\[
y = 4
\]

- Solution:
\[
\boxed{(-2, 4)}
\]

2. System:
\[
\begin{cases}
2x - 3y = 1 \\
3x + 2y = 12
\end{cases}
\]

- Step 1: Eliminate one variable by making coefficients equal.
- Multiply the first equation by 2 and the second equation by 3:
\[
\begin{cases}
4x - 6y = 2 \\
9x + 6y = 36
\end{cases}
\]

- Step 2: Add the equations to eliminate $y$.
\[
(4x - 6y) + (9x + 6y) = 2 + 36
\]
\[
13x = 38
\]
\[
x = \frac{38}{13}
\]

- Step 3: Solve for the other variable.
- Substitute $x = \frac{38}{13}$ into $2x - 3y = 1$:
\[
2\left(\frac{38}{13}\right) - 3y = 1
\]
\[
\frac{76}{13} - 3y = 1
\]
\[
\frac{76}{13} - 1 = 3y
\]
\[
\frac{76}{13} - \frac{13}{13} = 3y
\]
\[
\frac{63}{13} = 3y
\]
\[
y = \frac{21}{13}
\]

- Solution:
\[
\boxed{\left(\frac{38}{13}, \frac{21}{13}\right)}
\]

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#### Section 4: Solve Each of the Following Systems of Equations Using Any Method You Wish

1. System:
\[
\begin{cases}
\frac{3}{2}x + \frac{4}{3}y = -4 \\
2x - \frac{2}{3}y = -2
\end{cases}
\]

- Step 1: Eliminate fractions by multiplying through by the least common denominator.
- Multiply the first equation by 6:
\[
9x + 8y = -24
\]
- Multiply the second equation by 3:
\[
6x - 2y = -6
\]

- Step 2: Eliminate one variable.
- Multiply the second equation by 4 to align coefficients of $y$:
\[
\begin{cases}
9x + 8y = -24 \\
24x - 8y = -24
\end{cases}
\]
- Add the equations:
\[
(9x + 8y) + (24x - 8y) = -24 + (-24)
\]
\[
33x = -48
\]
\[
x = -\frac{48}{33} = -\frac{16}{11}
\]

- Step 3: Solve for the other variable.
- Substitute $x = -\frac{16}{11}$ into $6x - 2y = -6$:
\[
6\left(-\frac{16}{11}\right) - 2y = -6
\]
\[
-\frac{96}{11} - 2y = -6
\]
\[
-\frac{96}{11} + 6 = 2y
\]
\[
-\frac{96}{11} + \frac{66}{11} = 2y
\]
\[
-\frac{30}{11} = 2y
\]
\[
y = -\frac{15}{11}
\]

- Solution:
\[
\boxed{\left(-\frac{16}{11}, -\frac{15}{11}\right)}
\]

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Final Answers:

1. Graphing:
- (1): $\boxed{(3, 4)}$
- (2): $\boxed{(3, 1.5)}$
- (3): $\boxed{\text{No solution}}$

2. Substitution:
- (1): $\boxed{(5, 2)}$
- (2): $\boxed{\left(\frac{4}{7}, \frac{44}{7}\right)}$

3. Combination:
- (1): $\boxed{(-2, 4)}$
- (2): $\boxed{\left(\frac{38}{13}, \frac{21}{13}\right)}$

4. Any Method:
- (1): $\boxed{\left(-\frac{16}{11}, -\frac{15}{11}\right)}$