How do you write a system of equations with the solution (5,8)?

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How do you write a system of equations with the solution (5,8)?

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Answer 1 · 2018-03-19T21:24:15

There are many variations available. It is an extremely open question.

$y = \frac{1}{2} x + \frac{11}{2}$ $y=1/2x+11/2$
$y = - 2 x + 18$ $y=-2x+18$

Explanation:

$Comment$ $color(blue)("Comment")$

I am choosing to combine two straight line graphs.

If I used a quadratic there could be 2 solutions but the question definitely states THE solution (singular)

What you do is substitute the known common values for $(x, y) = (5, 8)$ $(x,y)=(5,8)$ and see what unfolds.
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$Consider the 1st straight line equation$ $color(blue)("Consider the 1st straight line equation")$

$y = m x + c$ $y=mx+c$

I choose to make $m = \frac{1}{2}$ $m=1/2$

$y = m x + c d \to d 8 = \frac{1}{2} (5) + c$ $y=mx+c color(white)("d") ->color(white)("d")8=1/2(5)+c$

Thus $c = 8 - \frac{5}{2} = \frac{11}{2}$ $c=8-5/2= 11/2$

So the equation of the straight line is $y = \frac{1}{2} x + \frac{11}{2}$ $color(lime)(y=1/2x+11/2)$
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$Consider the 2nd straight line equation$ $color(blue)("Consider the 2nd straight line equation")$

This time I choose to make the gradient negative (downward slope)

Set $m = - 2$ $m=-2$

$y = m x + c dd \to dd y = - 2 x + c dd \to dd 8 = - 2 (5) + c$ $y=mx+c color(white)("dd")->color(white)("dd")y=-2x+c color(white)("dd")->color(white)("dd")8=-2(5)+c$

From this $c = 8 + 10 = 18$ $c=8+10=18$ giving:

$y = - 2 x + 18$ $color(lime)(y=-2x+18)$

Tony B

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How do you write a system of equations with the solution (5,8)?

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