How do you solve the following system: $2 x - 4 y = 6, y + 4 x = 16$ $2x-4y=6 , y + 4x = 16$ ?

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How do you solve the following system: $2 x - 4 y = 6, y + 4 x = 16$ $2x-4y=6 , y + 4x = 16$ ?

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Answer 1 · 2018-05-17T03:15:23

See a solution process below:

Explanation:

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

$y + 4 x = 16$ $y + 4x = 16$

$y + 4 x - 4 x = 16 - 4 x$ $y + 4x - color(red)(4x) = 16 - color(red)(4x)$

$y + 0 = 16 - 4 x$ $y + 0 = 16 - 4x$

$y = 16 - 4 x$ $y = 16 - 4x$

Step 2) Substitute $(16 - 4 x)$ $(16 - 4x)$ for $y$ $y$ in the first equation and solve for $x$ $x$ :

$2 x - 4 y = 6$ $2x - 4y = 6$ becomes:

$2 x - 4 (16 - 4 x) = 6$ $2x - 4(16 - 4x) = 6$

$2 x - (4 \cdot 16) + (4 \cdot 4 x) = 6$ $2x - (4 * 16) + (4 * 4x) = 6$

$2 x - 64 + 16 x = 6$ $2x - 64 + 16x = 6$

$2 x + 16 x - 64 = 6$ $2x + 16x - 64 = 6$

$(2 + 16) x - 64 + 64 = 6 + 64$ $(2 + 16)x - 64 + color(red)(64) = 6 + color(red)(64)$

$18 x - 0 = 70$ $18x - 0 = 70$

$18 x = 70$ $18x = 70$

$\frac{18 x}{18} = \frac{70}{18}$ $(18x)/color(red)(18) = 70/color(red)(18)$

$x = \frac{35}{9}$ $x = 35/9$

Step 3) Substitute $\frac{35}{9}$ $35/9$ for $x$ $x$ in the solution to the second equation at the end of Step 1 and calculate $y$ $y$ :

$y = 16 - 4 x$ $y = 16 - 4x$ becomes:

$y = 16 - (4 \times \frac{35}{9})$ $y = 16 - (4 xx 35/9)$

$y = \frac{144}{9} - \frac{140}{9}$ $y = 144/9 - 140/9$

$y = \frac{144 - 140}{9}$ $y = (144 - 140)/9$

$y = \frac{4}{9}$ $y = 4/9$

The Solution Is:

$x = \frac{35}{9}$ $x = 35/9$ and $y = \frac{4}{9}$ $y = 4/9$

Or

$(\frac{35}{9}, \frac{4}{9})$ $(35/9, 4/9)$

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How do you solve the following system: $2 x - 4 y = 6, y + 4 x = 16$ $2x-4y=6 , y + 4x = 16$ ?

1 Answer

Explanation:

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How do you solve the following system: $2 x - 4 y = 6, y + 4 x = 16$ $2x-4y=6 , y + 4x = 16$ ?