Let's solve the system of linear equations given:
1) [tex]\( 5x + 3y = 21 \)[/tex]
2) [tex]\( -2x + 4y = 2 \)[/tex]
We will solve this system using the method of substitution or elimination.
### Step-by-Step Solution:
#### Step 1: Set up the equations in matrix form
We can represent the system of linear equations with a coefficient matrix [tex]\( A \)[/tex] and a constant matrix [tex]\( B \)[/tex]:
[tex]\[
A \cdot \begin{pmatrix} x \\ y \end{pmatrix} = B
\][/tex]
Where [tex]\( A = \begin{pmatrix} 5 & 3 \\ -2 & 4 \end{pmatrix} \)[/tex] and [tex]\( B = \begin{pmatrix} 21 \\ 2 \end{pmatrix} \)[/tex].
#### Step 2: Use substitution or elimination method
Using elimination:
1. Multiply Equation 1 by 2 to make the coefficients of [tex]\(x\)[/tex] in both equations equal:
[tex]\[
2 \cdot (5x + 3y) = 2 \cdot 21
\][/tex]
This yields:
[tex]\[
10x + 6y = 42 \quad \text{(Equation 3)}
\][/tex]
2. Add Equation 2 to Equation 3:
[tex]\[
(10x + 6y) + (-2x + 4y) = 42 + 2
\][/tex]
Simplifying this, we get:
[tex]\[
(10x - 2x) + (6y + 4y) = 44
\][/tex]
Which simplifies to:
[tex]\[
8x + 10y = 44 \quad \text{(Equation 4)}
\][/tex]
3. Now solve Equation 4 for [tex]\( y \)[/tex]:
[tex]\[
8x + 10y = 44
\][/tex]
Notice there might have been a likely approach mistake, instead, we normalize the equation first if needed. We can combine the coefficients accordingly instead.
Combine coefficients properly:
- Equation after normalization from the 4th:
Multiply Equation 2 by 5:
[tex]\[
5 \cdot (-2x + 4y) = 5 \cdot 2
\][/tex]
Yielding:
[tex]\[
- 10x + 20y = 10 \quad (Equation 5)
\][/tex]
Add results of valid Equation3 and Equation5:
10x+6y+(-10x+20y)=42+10
Simplifies to
26 y = 52
Recovered fast approach back. Valid substitution indeed says:
[tex]\[
y= 2
\][/tex]
#### Step 3: Solve for [tex]\(x\)[/tex]:
Using [tex]\(y=2\)[/tex] in first simpler equation
we combine coefficients as:
[tex]\[
5x + 3(2)=21\][/tex]
Solving:
5x +6=21
\]
So
\[
5x=15
obtaining result:
\[.
Thus the results through good solving steps, recover accurate results:
\boxed{ x = 3, y=2 }
Thus:
\boxed{(3,2)}
