Question

What is the solution to the system of equations below?

2 x minus y = 10 and y = negative one-half x + 5
(6, 2)
(6, –2)
(–6, –22)
(–6, 8)

Answers

There are no human answers yet.
Answered by GPT-4o mini
To solve the system of equations given:

1. \( 2x - y = 10 \)
2. \( y = -\frac{1}{2}x + 5 \)

We can use substitution or elimination. Here, we can substitute the second equation into the first equation.

From the second equation, we have:

\[ y = -\frac{1}{2}x + 5 \]

Now substitute this expression for \( y \) into the first equation:

\[
2x - \left(-\frac{1}{2}x + 5\right) = 10
\]

Simplifying this gives:

\[
2x + \frac{1}{2}x - 5 = 10
\]

Combining terms:

\[
2x + \frac{1}{2}x = \frac{4}{2}x + \frac{1}{2}x = \frac{5}{2}x
\]

So we have:

\[
\frac{5}{2}x - 5 = 10
\]

Now, add 5 to both sides:

\[
\frac{5}{2}x = 15
\]

Now, multiply both sides by \(\frac{2}{5}\):

\[
x = 15 \cdot \frac{2}{5} = 6
\]

Next, we substitute \( x = 6 \) back into the second equation to find \( y \):

\[
y = -\frac{1}{2}(6) + 5
\]
\[
y = -3 + 5 = 2
\]

So, the solution to the system of equations is:

\[
(x, y) = (6, 2)
\]

Therefore, the answer is \((6, 2)\).

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