Question: 3) Solve each linear system using the method of ELIMINATION [9]
a) ℓ₁: 3x + 4y = 4
ℓ₂: -2x - 4y = 8
Solution:
b) ℓ₁: 3x + 5y = 1
ℓ₂: 4x + y = 7
Solution:
c) ℓ₁: 2x - 5y = 2
ℓ₂: -5x = -3y + 2
Solution:
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1. **Problem:** Solve each linear system using **elimination**.
2. **a) Solve the system**
$$3x + 4y = 4$$
$$-2x - 4y = 8$$
**Rule:** In elimination, we add or subtract equations to remove one variable.
Add the equations:
$$\begin{aligned}
(3x + 4y) + (-2x - 4y) &= 4 + 8
\end{aligned}$$
$$\begin{aligned}
x &= 12
\end{aligned}$$
Substitute $x = 12$ into $3x + 4y = 4$:
$$\begin{aligned}
3(12) + 4y &= 4
\end{aligned}$$
$$\begin{aligned}
36 + 4y &= 4
\end{aligned}$$
$$\begin{aligned}
4y &= 4 - 36
\end{aligned}$$
$$\begin{aligned}
4y &= -32
\end{aligned}$$
$$\begin{aligned}
y &= \frac{-32}{4}
\end{aligned}$$
$$\begin{aligned}
\cancel{4}y &= \frac{\cancel{-32}}{\cancel{4}}
\end{aligned}$$
$$\begin{aligned}
y &= -8
\end{aligned}$$
**Answer for a:** $x = 12$, $y = -8$.
3. **b) Solve the system**
$$3x + 5y = 1$$
$$4x + y = 7$$
Multiply the second equation by $-5$ so the $y$-terms cancel:
$$\begin{aligned}
-5(4x + y) &= -5(7)
\end{aligned}$$
$$\begin{aligned}
-20x - 5y &= -35
\end{aligned}$$
Now add with the first equation:
$$\begin{aligned}
(3x + 5y) + (-20x - 5y) &= 1 + (-35)
\end{aligned}$$
$$\begin{aligned}
-17x &= -34
\end{aligned}$$
$$\begin{aligned}
x &= \frac{-34}{-17}
\end{aligned}$$
$$\begin{aligned}
\cancel{-17}x &= \frac{\cancel{-34}}{\cancel{-17}}
\end{aligned}$$
$$\begin{aligned}
x &= 2
\end{aligned}$$
Substitute $x = 2$ into $4x + y = 7$:
$$\begin{aligned}
4(2) + y &= 7
\end{aligned}$$
$$\begin{aligned}
8 + y &= 7
\end{aligned}$$
$$\begin{aligned}
y &= 7 - 8
\end{aligned}$$
$$\begin{aligned}
y &= -1
\end{aligned}$$
**Answer for b:** $x = 2$, $y = -1$.
4. **c) Solve the system**
$$2x - 5y = 2$$
$$-5x = -3y + 2$$
First rewrite the second equation in standard form:
$$\begin{aligned}
-5x + 3y &= 2
\end{aligned}$$
Multiply the first equation by $3$ and the second equation by $5$ so the $y$-terms cancel:
$$\begin{aligned}
3(2x - 5y) &= 3(2)
\end{aligned}$$
$$\begin{aligned}
6x - 15y &= 6
\end{aligned}$$
$$\begin{aligned}
5(-5x + 3y) &= 5(2)
\end{aligned}$$
$$\begin{aligned}
-25x + 15y &= 10
\end{aligned}$$
Add the new equations:
$$\begin{aligned}
(6x - 15y) + (-25x + 15y) &= 6 + 10
\end{aligned}$$
$$\begin{aligned}
-19x &= 16
\end{aligned}$$
$$\begin{aligned}
x &= \frac{16}{-19}
\end{aligned}$$
$$\begin{aligned}
\cancel{-19}x &= \frac{\cancel{16}}{\cancel{-19}}
\end{aligned}$$
$$\begin{aligned}
x &= -\frac{16}{19}
\end{aligned}$$
Substitute into $2x - 5y = 2$:
$$\begin{aligned}
2\left(-\frac{16}{19}\right) - 5y &= 2
\end{aligned}$$
$$\begin{aligned}
-\frac{32}{19} - 5y &= 2
\end{aligned}$$
$$\begin{aligned}
-5y &= 2 + \frac{32}{19}
\end{aligned}$$
$$\begin{aligned}
-5y &= \frac{38}{19} + \frac{32}{19}
\end{aligned}$$
$$\begin{aligned}
-5y &= \frac{70}{19}
\end{aligned}$$
$$\begin{aligned}
y &= \frac{70}{19} \div (-5)
\end{aligned}$$
$$\begin{aligned}
y &= \frac{70}{19} \cdot \frac{-1}{5}
\end{aligned}$$
$$\begin{aligned}
y &= -\frac{14}{19}
\end{aligned}$$
**Answer for c:** $x = -\frac{16}{19}$, $y = -\frac{14}{19}$.
5. **Final answers:**
a) $\left(12, -8\right)$
b) $\left(2, -1\right)$
c) $\left(-\frac{16}{19}, -\frac{14}{19}\right)$