EXERCISE 4 Solve for $$ x $$ and $$ y $$ by using the method of elimination: $$ \begin{array}{ll} ext { (a) } x-y=3 ext { and } 2 x+y=9 & ext { (b) } x+y=-6 ext { and } 3 x+y=-10 \ ext { (c) } x+2 y=5 ext { and } x-y=-1 & ext { (d) } 3 x+5 y=8 ext { and } x-2 y=-1 \ ext { (e) } 2 x-3 y=10 ext { and } 4 x+5 y=42 & ext { (f) } 3 y-4 x=7 ext { and } 2 x+5 y=16\end{array} $$

Question

EXERCISE 4 Solve for $$ x $$ and $$ y $$ by using the method of elimination: $$ \begin{array}{ll}	ext { (a) } x-y=3 	ext { and } 2 x+y=9 & 	ext { (b) } x+y=-6 	ext { and } 3 x+y=-10 \ 	ext { (c) } x+2 y=5 	ext { and } x-y=-1 & 	ext { (d) } 3 x+5 y=8 	ext { and } x-2 y=-1 \ 	ext { (e) } 2 x-3 y=10 	ext { and } 4 x+5 y=42 & 	ext { (f) } 3 y-4 x=7 	ext { and } 2 x+5 y=16\end{array} $$

UpStudy AI · Accepted Answer

**(a)**  
We are given  
$$
x - y = 3 \quad \text{and} \quad 2x + y = 9.
$$  
*Step 1: Add the two equations to eliminate $$ y $$:*  
$$
(x - y) + (2x + y) = 3 + 9 \quad \Longrightarrow \quad 3x = 12.
$$  
*Step 2: Solve for $$ x $$:*  
$$
x = \frac{12}{3} = 4.
$$  
*Step 3: Substitute $$ x = 4 $$ into the first equation to solve for $$ y $$:*  
$$
4 - y = 3 \quad \Longrightarrow \quad -y = -1 \quad \Longrightarrow \quad y = 1.
$$

---

**(b)**  
We are given  
$$
x + y = -6 \quad \text{and} \quad 3x + y = -10.
$$  
*Step 1: Subtract the first equation from the second to eliminate $$ y $$:*  
$$
(3x + y) - (x + y) = -10 - (-6) \quad \Longrightarrow \quad 2x = -4.
$$  
*Step 2: Solve for $$ x $$:*  
$$
x = \frac{-4}{2} = -2.
$$  
*Step 3: Substitute $$ x = -2 $$ into the first equation to solve for $$ y $$:*  
$$
-2 + y = -6 \quad \Longrightarrow \quad y = -6 + 2 = -4.
$$

---

**(c)**  
We are given  
$$
x + 2y = 5 \quad \text{and} \quad x - y = -1.
$$  
*Step 1: Subtract the second equation from the first to eliminate $$ x $$:*  
$$
(x + 2y) - (x - y) = 5 - (-1) \quad \Longrightarrow \quad 3y = 6.
$$  
*Step 2: Solve for $$ y $$:*  
$$
y = \frac{6}{3} = 2.
$$  
*Step 3: Substitute $$ y = 2 $$ into the second equation to solve for $$ x $$:*  
$$
x - 2 = -1 \quad \Longrightarrow \quad x = -1 + 2 = 1.
$$

---

**(d)**  
We are given  
$$
3x + 5y = 8 \quad \text{and} \quad x - 2y = -1.
$$  
*Step 1: Multiply the second equation by 3 to make the $$ x $$-coefficients the same:*  
$$
3(x - 2y) = 3(-1) \quad \Longrightarrow \quad 3x - 6y = -3.
$$  
*Step 2: Subtract the new equation from the first equation to eliminate $$ x $$:*  
$$
(3x + 5y) - (3x - 6y) = 8 - (-3) \quad \Longrightarrow \quad 11y = 11.
$$  
*Step 3: Solve for $$ y $$:*  
$$
y = \frac{11}{11} = 1.
$$  
*Step 4: Substitute $$ y = 1 $$ into the second original equation to solve for $$ x $$:*  
$$
x - 2(1) = -1 \quad \Longrightarrow \quad x - 2 = -1 \quad \Longrightarrow \quad x = 1.
$$

---

**(e)**  
We are given  
$$
2x - 3y = 10 \quad \text{and} \quad 4x + 5y = 42.
$$  
*Step 1: Multiply the first equation by 2 to align the $$ x $$-coefficients with the second equation:*  
$$
2(2x - 3y) = 2(10) \quad \Longrightarrow \quad 4x - 6y = 20.
$$  
*Step 2: Subtract this new equation from the second equation to eliminate $$ x $$:*  
$$
(4x + 5y) - (4x - 6y) = 42 - 20 \quad \Longrightarrow \quad 11y = 22.
$$  
*Step 3: Solve for $$ y $$:*  
$$
y = \frac{22}{11} = 2.
$$  
*Step 4: Substitute $$ y = 2 $$ into the first equation to solve for $$ x $$:*  
$$
2x - 3(2) = 10 \quad \Longrightarrow \quad 2x - 6 = 10 \quad \Longrightarrow \quad 2x = 16 \quad \Longrightarrow \quad x = 8.
$$

---

**(f)**  
We are given  
$$
3y - 4x = 7 \quad \text{and} \quad 2x + 5y = 16.
$$  
*Step 1: Rewrite the first equation in standard form:*  
$$
-4x + 3y = 7.
$$  
*Step 2: Multiply the second equation by 2 to facilitate eliminating $$ x $$:*  
$$
2(2x + 5y) = 2(16) \quad \Longrightarrow \quad 4x + 10y = 32.
$$  
*Step 3: Add the modified equations to eliminate $$ x $$:*  
$$
(-4x + 3y) + (4x + 10y) = 7 + 32 \quad \Longrightarrow \quad 13y = 39.
$$
*Step 4: Solve for $$ y $$:*  
$$
y = \frac{39}{13} = 3.
$$
*Step 5: Substitute $$ y = 3 $$ into the second original equation to solve for $$ x $$:*  
$$
2x + 5(3) = 16 \quad \Longrightarrow \quad 2x + 15 = 16 \quad \Longrightarrow \quad 2x = 1 \quad \Longrightarrow \quad x = \frac{1}{2}.
$$
**(a)** $$ x = 4 $$, $$ y = 1 $$ **(b)** $$ x = -2 $$, $$ y = -4 $$ **(c)** $$ x = 1 $$, $$ y = 2 $$ **(d)** $$ x = 1 $$, $$ y = 1 $$ **(e)** $$ x = 8 $$, $$ y = 2 $$ **(f)** $$ x = \frac{1}{2} $$, $$ y = 3 $$

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