How do you solve the following system: $x - 4 y = 2, 5 x + 2 y = 20$ $x-4y=2, 5x+2y=20$ ?

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How do you solve the following system: $x - 4 y = 2, 5 x + 2 y = 20$ $x-4y=2, 5x+2y=20$ ?

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Answer 1 · 2016-01-18T18:48:22

$x = \frac{42}{11}, y = \frac{5}{11}$ $x=42/11,y=5/11$

Explanation:

Take the first equation.

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

Add $4 y$ $4y$ to both sides.

$x = 4 y + 2$ $x=4y+2$

Since we know that $x = 4 y + 2$ $x=4y+2$ , we can place $x$ $x$ with $4 y + 2$ $4y+2$ in the other equation.

$5 x + 2 y = 20$ $5color(red)x+2y=20$

$5 (4 y + 2) + 2 y = 20$ $5color(red)((4y+2))+2y=20$

Distribute and solve for $y$ $y$ .

$20 y + 10 + 2 y = 20$ $20y+10+2y=20$

$22 y = 10$ $22y=10$

$y = \frac{10}{22} = \frac{5}{11}$ $y=10/22=5/11$

Now that we know the value for $y$ $y$ , plug it into either equation to solve for $x$ $x$ . I'm choosing the first equation since it's simpler, but the same answer would come from plugging the value into the second equation.

$x - 4 y = 2$ $x-4color(blue)y=2$

$x - 4 (\frac{5}{11}) = 2$ $x-4color(blue)((5/11))=2$

$x - \frac{20}{11} = 2$ $x-20/11=2$

$x = \frac{22}{11} + \frac{20}{11} = \frac{42}{11}$ $x=22/11+20/11=42/11$

Since $x = \frac{42}{11}$ $x=42/11$ and $y = \frac{5}{11}$ $y=5/11$ , the two lines should intersect at the point $(\frac{42}{11}, \frac{5}{11})$ $(42/11,5/11)$ or $(3.818, 0.455)$ $(3.818,0.455)$ .

We can check a graph:

graph{(x-4y-2)(5x+2y-20)=0 [-5.26, 10.54, -3.38, 4.516]}

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How do you solve the following system: $x - 4 y = 2, 5 x + 2 y = 20$ $x-4y=2, 5x+2y=20$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the following system: x−4y=2,5x+2y=20x-4y=2, 5x+2y=20 ?

1 Answer

Explanation:

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Impact of this question

How do you solve the following system: $x - 4 y = 2, 5 x + 2 y = 20$ $x-4y=2, 5x+2y=20$ ?