To solve the system of linear equations given by
[tex]\[4x - 3y = -9\][/tex]
[tex]\[x + 3y = 4\][/tex]
we will follow a step-by-step procedure to find the value of [tex]\(y\)[/tex].
Step 1: Write down the equations.
We start with:
[tex]\[4x - 3y = -9 \quad ...(1)\][/tex]
[tex]\[x + 3y = 4 \quad ...(2)\][/tex]
Step 2: Solve one equation for one variable.
We can first solve Equation (2) for [tex]\(x\)[/tex]:
[tex]\[x + 3y = 4\][/tex]
So,
[tex]\[x = 4 - 3y \quad ...(3)\][/tex]
Step 3: Substitute the expression obtained into the other equation.
We substitute [tex]\(x = 4 - 3y\)[/tex] from Equation (3) into Equation (1):
[tex]\[4(4 - 3y) - 3y = -9\][/tex]
Step 4: Simplify and solve for [tex]\(y\)[/tex].
Expand and simplify:
[tex]\[16 - 12y - 3y = -9\][/tex]
[tex]\[16 - 15y = -9\][/tex]
Move constants to one side:
[tex]\[16 + 9 = 15y\][/tex]
[tex]\[25 = 15y\][/tex]
Divide by 15:
[tex]\[y = \frac{25}{15}\][/tex]
[tex]\[y = \frac{5}{3}\][/tex]
Step 5: Verify the solution.
It’s helpful to check the solution by substituting [tex]\(y = \frac{5}{3}\)[/tex] back into the second equation to find [tex]\(x\)[/tex]:
[tex]\[x + 3 \left(\frac{5}{3}\right) = 4\][/tex]
[tex]\[ x + 5 = 4 \][/tex]
[tex]\[ x = -1 \][/tex]
Both values satisfy the original equations:
- Substituting into the first equation:
[tex]\[ 4(-1) - 3\left(\frac{5}{3}\right) = -4 - 5 = -9 \][/tex]
- Substituting into the second equation:
[tex]\[ -1 + 3\left(\frac{5}{3}\right) = -1 + 5 = 4 \][/tex]
Thus, the value of [tex]\(y\)[/tex] is correctly found to be [tex]\(\frac{5}{3}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{E} \][/tex]
