How do you solve the following linear system: $x + y = 4, - 3 x + y = - 8$ $x+y=4, -3x+y=-8$ ?

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How do you solve the following linear system: $x + y = 4, - 3 x + y = - 8$ $x+y=4, -3x+y=-8$ ?

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Answer 1 · 2018-05-31T12:19:45

See a solution process below:

Explanation:

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

$x + y = 4$ $x + y = 4$

$x + y - y = 4 - y$ $x + y - color(red)(y) = 4 - color(red)(y)$

$x + 0 = 4 - y$ $x + 0 = 4 - y$

$x = 4 - y$ $x = 4 - y$

Step 2) Substitute $(4 - y)$ $(4 - y)$ for $x$ $x$ in the second equation and solve for $y$ $y$ :

$- 3 x + y = - 8$ $-3x + y = -8$ becomes:

$- 3 (4 - y) + y = - 8$ $-3(4 - y) + y = -8$

$(- 3 \times 4) + (3 \times y) + y = - 8$ $(-3 xx 4) + (3 xx y) + y = -8$

$- 12 + 3 y + y = - 8$ $-12 + 3y + y = -8$

$- 12 + 12 + 3 y + y = - 8 + 12$ $-12 + color(red)(12) + 3y + y = -8 + color(red)(12)$

$0 + 3 y + y = 4$ $0 + 3y + y = 4$

$3 y + y = 4$ $3y + y = 4$

$3 y + 1 y = 4$ $3y + 1y = 4$

$(3 + 1) y = 4$ $(3 + 1)y = 4$

$4 y = 4$ $4y = 4$

$\frac{4 y}{4} = \frac{4}{4}$ $(4y)/color(red)(4) = 4/color(red)(4)$

$y = 1$ $y = 1$

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

$x = 4 - y$ $x = 4 - y$ becomes:

$x = 4 - 1$ $x = 4 - 1$

$x = 3$ $x = 3$

The Solution Is:

$x = 3$ $x = 3$ and $y = 1$ $y = 1$

Or

$(3, 1)$ $(3, 1)$

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How do you solve the following linear system: $x + y = 4, - 3 x + y = - 8$ $x+y=4, -3x+y=-8$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the following linear system: x+y=4,−3x+y=−8x+y=4, -3x+y=-8 ?

1 Answer

Explanation:

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Impact of this question

How do you solve the following linear system: $x + y = 4, - 3 x + y = - 8$ $x+y=4, -3x+y=-8$ ?