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]
