How do you solve the system of equations $2 x + y = 8$ $2x + y = 8$ and $3 x - y = 7$ $3x - y = 7$ ?

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How do you solve the system of equations $2 x + y = 8$ $2x + y = 8$ and $3 x - y = 7$ $3x - y = 7$ ?

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Answer 1 · 2018-04-11T13:10:51

See a solution process below:

Explanation:

Step 1) Solve the first equation for $y$ $y$ :

$2 x + y = 8$ $2x + y = 8$

$2 x - 2 x + y = 8 - 2 x$ $2x - color(red)(2x) + y = 8 - color(red)(2x)$

$0 + y = 8 - 2 x$ $0 + y = 8 - 2x$

$y = 8 - 2 x$ $y = 8 - 2x$

Step 2) Substitute $(8 - 2 x)$ $(8 - 2x)$ for $y$ $y$ in the second equation and solve for $x$ $x$ :

$3 x - y = 7$ $3x - y = 7$ becomes:

$3 x - (8 - 2 x) = 7$ $3x - (8 - 2x) = 7$

$3 x - 8 + 2 x = 7$ $3x - 8 + 2x = 7$

$3 x - 8 + 8 + 2 x = 7 + 8$ $3x - 8 + color(red)(8) + 2x = 7 + color(red)(8)$

$3 x - 0 + 2 x = 15$ $3x - 0 + 2x = 15$

$3 x + 2 x = 15$ $3x + 2x = 15$

$(3 + 2) x = 15$ $(3 + 2)x = 15$

$5 x = 15$ $5x = 15$

$\frac{5 x}{5} = \frac{15}{5}$ $(5x)/color(red)(5) = 15/color(red)(5)$

$x = 3$ $x = 3$

Step 3) Substitute $3$ $3$ for $x$ $x$ in the solution to the first equation at the end of Step 1 and calculate $y$ $y$ :

$y = 8 - 2 x$ $y = 8 - 2x$ becomes:

$y = 8 - (2 \times 3)$ $y = 8 - (2 xx 3)$

$y = 8 - 6$ $y = 8 - 6$

$y = 2$ $y = 2$

**The Solution Is:

$x = 3$ $x = 3$ and $y = 2$ $y = 2$

Or

$(3, 2)$ $(3, 2)$

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How do you solve the system of equations $2 x + y = 8$ $2x + y = 8$ and $3 x - y = 7$ $3x - y = 7$ ?

1 Answer

Explanation:

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