Answer :

Sure! To solve this system of equations using the substitution method, follow these steps:

1. Given System of Equations:
[tex]\[ \begin{cases} 5x + 2y = 1 & \quad \text{(Equation 1)} \\ -3x + 3y = 5 & \quad \text{(Equation 2)} \end{cases} \][/tex]

2. Solve Equation 2 for one of the variables: Let's solve for \( y \) in terms of \( x \):

[tex]\[ -3x + 3y = 5 \][/tex]

Add \( 3x \) to both sides:

[tex]\[ 3y = 3x + 5 \][/tex]

Divide both sides by 3:

[tex]\[ y = x + \frac{5}{3} \][/tex]

3. Substitute \( y \) in Equation 1: Now that we have \( y \) in terms of \( x \), substitute it into Equation 1:

[tex]\[ 5x + 2\left(x + \frac{5}{3}\right) = 1 \][/tex]

4. Simplify and Solve for \( x \):

Distribute the 2:
[tex]\[ 5x + 2x + \frac{10}{3} = 1 \][/tex]

Combine like terms:
[tex]\[ 7x + \frac{10}{3} = 1 \][/tex]

To clear the fraction, multiply the entire equation by 3:
[tex]\[ 21x + 10 = 3 \][/tex]

Subtract 10 from both sides:
[tex]\[ 21x = -7 \][/tex]

Divide both sides by 21:
[tex]\[ x = -\frac{1}{3} \][/tex]

5. Substitute \( x \) back into the expression for \( y \):

We previously found \( y = x + \frac{5}{3} \). Substitute \( x = -\frac{1}{3} \):

[tex]\[ y = -\frac{1}{3} + \frac{5}{3} \][/tex]

Combine the fractions:
[tex]\[ y = \frac{-1 + 5}{3} = \frac{4}{3} \][/tex]

Therefore, the solution for the system of equations is:

[tex]\[ \boxed{\left( -\frac{1}{3}, \frac{4}{3} \right)} \][/tex]

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