How do you solve the following system: $3 x + y = 8, 2 x - y = 7$ $3x+y=8, 2x-y=7$ ?

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How do you solve the following system: $3 x + y = 8, 2 x - y = 7$ $3x+y=8, 2x-y=7$ ?

1 Answer

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Answer 1 · 2018-04-18T06:01:58

$x = 3$ $x=3$
$y = - 1$ $y=-1$

Explanation:

Given -

$3 x + y = 8$ $3x+y=8$ ------------- (1)
$2 x - y = 7$ $2x-y=7$ ---------------(2) ----- $(1) + (2)$ $(1) + (2)$

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

$x = \frac{15}{5} = 3$ $x=15/5=3$

$x = 3$ $x=3$

Substitute > $x = 3$ $x=3$ in equation (1)

$3 (3) + y = 8$ $3(3)+y=8$

$9 + y = 8$ $9+y=8$

$y = 8 - 9 = - 1$ $y=8-9=-1$

$y = - 1$ $y=-1$

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How do you solve the following system: $3 x + y = 8, 2 x - y = 7$ $3x+y=8, 2x-y=7$ ?

1 Answer

Explanation:

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How do you solve the following system: $3 x + y = 8, 2 x - y = 7$ $3x+y=8, 2x-y=7$ ?