Question

What is the y value of the solution to the system of equations shown below?
4x+3y=10
3x-2y=5


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Answer 1

To solve the system of equations:

1. \( 4x + 3y = 10 \) (Equation 1)
2. \( 3x - 2y = 5 \) (Equation 2)

We can use substitution or elimination. Here, I’ll use the elimination method.

First, we can try to eliminate one variable. Let's eliminate \(y\). To do that, we can multiply Equation 1 by 2 and Equation 2 by 3 to make the coefficients of \(y\) the same:

\[
2(4x + 3y) = 2(10) \implies 8x + 6y = 20 \quad \text{(Equation 3)}
\]
\[
3(3x - 2y) = 3(5) \implies 9x - 6y = 15 \quad \text{(Equation 4)}
\]

Now, we can add Equation 3 and Equation 4:

\[
(8x + 6y) + (9x - 6y) = 20 + 15
\]
\[
8x + 9x + 6y - 6y = 35
\]
\[
17x = 35
\]
\[
x = \frac{35}{17}
\]

Now, substitute \(x\) back into one of the original equations to find \(y\). Let's use Equation 1:

\[
4\left(\frac{35}{17}\right) + 3y = 10
\]
\[
\frac{140}{17} + 3y = 10
\]

Convert \(10\) into a fraction with a denominator of \(17\):

\[
10 = \frac{170}{17}
\]

So we have:

\[
\frac{140}{17} + 3y = \frac{170}{17}
\]

Subtract \(\frac{140}{17}\) from both sides:

\[
3y = \frac{170}{17} - \frac{140}{17}
\]
\[
3y = \frac{30}{17}
\]

Now, divide by \(3\):

\[
y = \frac{30}{17 \cdot 3} = \frac{30}{51} = \frac{10}{17}
\]

Thus, the value of \(y\) in the solution to the system of equations is:

\[
\boxed{\frac{10}{17}}
\]