7. solve each system of equations.
a. $\begin{cases} 2x + y = 13 \\ 10x - 3y = 1 \end{cases}$
b. $\begin{cases} -4x + \frac{1}{2}y = 25 \\ 8x + 3y = 46 \end{cases}$

Question

Accepted Answer

### Part a
# Explanation:
## Step1: Solve for $ y $ from the first equation
From $ 2x + y = 13 $, we get $ y = 13 - 2x $.
## Step2: Substitute $ y $ into the second equation
Substitute $ y = 13 - 2x $ into $ 10x - 3y = 1 $:
$$
\begin{align*}
10x - 3(13 - 2x) &= 1 \
10x - 39 + 6x &= 1 \
16x - 39 &= 1 \
16x &= 1 + 39 \
16x &= 40 \
x &= \frac{40}{16} \
x &= 2.5
\end{align*}
$$
## Step3: Find $ y $ using $ x = 2.5 $
Substitute $ x = 2.5 $ into $ y = 13 - 2x $:
$$
\begin{align*}
y &= 13 - 2(2.5) \
y &= 13 - 5 \
y &= 8
\end{align*}
$$
# Answer:
$ x = 2.5 $, $ y = 8 $

### Part b
# Explanation:
## Step1: Eliminate $ x $ by multiplying the first equation
Multiply $ -4x + \frac{1}{2}y = 25 $ by 2 to get $ -8x + y = 50 $.
## Step2: Add the new equation to the second equation
Add $ -8x + y = 50 $ and $ 8x + 3y = 46 $:
$$
\begin{align*}
(-8x + y) + (8x + 3y) &= 50 + 46 \
4y &= 96 \
y &= \frac{96}{4} \
y &= 24
\end{align*}
$$
## Step3: Find $ x $ using $ y = 24 $
Substitute $ y = 24 $ into $ -4x + \frac{1}{2}y = 25 $:
$$
\begin{align*}
-4x + \frac{1}{2}(24) &= 25 \
-4x + 12 &= 25 \
-4x &= 25 - 12 \
-4x &= 13 \
x &= -\frac{13}{4} \
x &= -3.25
\end{align*}
$$
# Answer:
$ x = -3.25 $, $ y = 24 $

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Solve for \( y \) from the first equation

Step2: Substitute \( y \) into the second equation

Step3: Find \( y \) using \( x = 2.5 \)

Step1: Eliminate \( x \) by multiplying the first equation

Step2: Add the new equation to the second equation

Step3: Find \( x \) using \( y = 24 \)

Answer:

Part b