Question

1. solve each system of equations.

a. \\(\

$$\begin{cases} 2x + y = 13 \\\\ 10x - 3y = 1 \\end{cases}$$

\\)
b. \\(\

$$\begin{cases} -4x + \\frac{1}{2}y = 25 \\\\ 8x + 3y = 46 \\end{cases}$$

\\)

Explanation:

Part a

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 \( y = 13 - 2x \):
\[

$$\begin{align*} y &= 13 - 2(2.5) \\ y &= 13 - 5 \\ y &= 8 \end{align*}$$

\]

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:

\( x = 2.5 \), \( y = 8 \)

Part b