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)

1 answer

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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