Practice Problems

Problems:

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Problem 1

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Problem 2

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Problem 3

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Solutions:

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Problem 1:

Step 1: Adjust the equations to eliminate one variable.

• We can eliminate $y$ by multiplying Equation B by $-3$, which will allow the $y$ terms to cancel out when added to Equation $A$. Multiplying Equation $B$ by $-3$: $-6x + 3y = -30 \quad \text{(Equation C)}$

Step 2: Add the equations to eliminate $y$.

• Add Equation $A$ and Equation $C$: $(2x - 3y) + (-6x + 3y) = 14 + (-30)$ Simplifying: $-4x = -16$ Step 3: Solve for $x$.

• Divide both sides by $-4$: $x = \frac{-16}{-4} = 4$ Step 4: Substitute $x = 4$ back into one of the original equations to solve for $y$.

• Substitute into Equation $B$: $2(4) - y = 10$ Simplifying: $8 - y = 10$

• Subtract $8$ from both sides: $-y = 2$

• Multiply both sides by $-1$: $y = -2$ Solution: The solution is $x = 4$ and $y = -2$.

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Problem 2:

Step 1: Adjust the equations to eliminate one variable.

• We can eliminate $y$ by multiplying Equation $A$ by $4$, which will allow the $y$ terms to cancel out when added to Equation $B$. Multiplying Equation $A$ by $4$: $12x + 4y = 20 \quad \text{(Equation C)}$

Step 2: Add the equations to eliminate $y$.

• Add Equation $C$ and Equation $B$: $(12x + 4y) + (5x - 4y) = 20 + (-3)$ Simplifying: $17x = 17$ Step 3: Solve for $x$.

• Divide both sides by $17$: $x = \frac{17}{17} = 1$ Step 4: Substitute $x = 1$ back into one of the original equations to solve for $y$.

• Substitute into Equation $A$: $3(1) + y = 5$ Simplifying: $3 + y = 5$

• Subtract $3$ from both sides: $y = 2$ Solution: The solution is $x = 1$ and $y = 2$.

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Problem 3:

Step 1: Adjust the equations to eliminate one variable.

• We can eliminate $y$ by multiplying Equation $A$ by $2$, which will allow the $y$ terms to cancel out when added to Equation $B$. Multiplying Equation $A$ by $2$: $4x - 2y = 24 \quad \text{(Equation C)}$

Step 2: Add the equations to eliminate $y$.

• Add Equation $C$ and Equation $B$: $(4x - 2y) + (3x + 2y) = 24 + 11$ Simplifying: $7x = 35$ Step 3: Solve for $x$.

• Divide both sides by $7$: $x = \frac{35}{7} = 5$ Step 4: Substitute $x = 5$ back into one of the original equations to solve for $y$.

• Substitute into Equation $A$: $2(5) - y = 12$ Simplifying: $10 - y = 12$

• Subtract $10$ from both sides: $-y = 2$

• Multiply both sides by $-1$: $y = -2$ Solution: The solution is $x = 5$ and $y = -2$.

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