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Solve the following system of equations:

[tex]\[
\begin{aligned}
\text{(1) } 5x + 3y &= 21 \\
\text{(2) } -2x + 4y &= 2
\end{aligned}
\][/tex]

Sagot :

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)}