How do you solve the system $y = 4 x - 1$ $y=4x-1$ and $y = 2 x - 5$ $y=2x-5$ using substitution?

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How do you solve the system $y = 4 x - 1$ $y=4x-1$ and $y = 2 x - 5$ $y=2x-5$ using substitution?

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Answer 1 · 2017-08-01T22:04:07

See a solution process below:

Explanation:

Step 1) Because the first equation is already solved for $y$ $y$ we can substitute $(4 x - 1)$ $(4x - 1)$ for $y$ $y$ in the second equation and solve for $x$ $x$ :

$4 x - 1 = 2 x - 5$ $4x - 1 = 2x - 5$

$- 2 x + 4 x - 1 + 1 = - 2 x + 2 x - 5 + 1$ $-color(blue)(2x) + 4x - 1 + color(red)(1) = -color(blue)(2x) + 2x - 5 + color(red)(1)$

$(- 2 + 4) x - 0 = 0 - 4$ $(-color(blue)(2) + 4)x - 0 = 0 - 4$

$2 x = - 4$ $2x = -4$

$\frac{2 x}{2} = - \frac{4}{2}$ $(2x)/color(red)(2) = -4/color(red)(2)$

$(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -2$

$x = -2$

Step 2) Substitute $-2$ for $x$ in the first equation and calculate $y$ :

$y = 4x - 1$ becomes:

$y = (4 xx -2) - 1$

$y = -8 - 1$

$y = -9$

The Solution Is: $x = -2$ and $y = -9$ or $(-2, -9)$

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How do you solve the system $y = 4 x - 1$ $y=4x-1$ and $y = 2 x - 5$ $y=2x-5$ using substitution?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the system y=4x-1y=4x−1y=4x-1 and y=2x-5y=2x−5y=2x-5 using substitution?

1 Answer

Explanation:

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Impact of this question

How do you solve the system $y = 4 x - 1$ $y=4x-1$ and $y = 2 x - 5$ $y=2x-5$ using substitution?