To solve the system of equations:
[tex]\[
\begin{cases}
3x + 2y = 16 \\
5x - 3y = 14
\end{cases}
\][/tex]
we'll follow a step-by-step algebraic method to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### Step 1: Write Down the Equations
We have the two linear equations:
[tex]\[
3x + 2y = 16 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
5x - 3y = 14 \quad \text{(Equation 2)}
\][/tex]
### Step 2: Prepare to Eliminate One Variable
We'll first eliminate [tex]\( y \)[/tex]. To do this, we'll make the coefficients of [tex]\( y \)[/tex] in both equations the same by finding a common multiple. The least common multiple of 2 and 3 is 6. Therefore:
Multiply Equation 1 by 3:
[tex]\[
3(3x + 2y) = 3(16) \Rightarrow 9x + 6y = 48 \quad \text{(Equation 3)}
\][/tex]
Multiply Equation 2 by 2:
[tex]\[
2(5x - 3y) = 2(14) \Rightarrow 10x - 6y = 28 \quad \text{(Equation 4)}
\][/tex]
### Step 3: Add the Equations to Eliminate [tex]\( y \)[/tex]
Now, we'll add Equation 3 and Equation 4:
[tex]\[
(9x + 6y) + (10x - 6y) = 48 + 28
\][/tex]
This simplifies to:
[tex]\[
9x + 10x = 76
\][/tex]
[tex]\[
19x = 76
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{76}{19} = 4
\][/tex]
### Step 4: Substitute [tex]\( x = 4 \)[/tex] into One of the Original Equations
We'll use [tex]\( x = 4 \)[/tex] in Equation 1 to find [tex]\( y \)[/tex]:
[tex]\[
3(4) + 2y = 16
\][/tex]
[tex]\[
12 + 2y = 16
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
2y = 16 - 12
\][/tex]
[tex]\[
2y = 4
\][/tex]
[tex]\[
y = \frac{4}{2} = 2
\][/tex]
### Step 5: State the Solution
The solution to the system of equations is:
[tex]\[
(x, y) = (4, 2)
\][/tex]
This means the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations are [tex]\( x = 4 \)[/tex] and [tex]\( y = 2 \)[/tex].
