How do you solve the system $y = x + 7$ $y=x+7$ , $y = {(x + 3)}^{2} - 8$ $y=(x+3)^2-8$ ?

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How do you solve the system $y = x + 7$ $y=x+7$ , $y = {(x + 3)}^{2} - 8$ $y=(x+3)^2-8$ ?

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Answer 1 · 2015-11-15T14:08:27

There are two solutions x=1, y=8 and x=-6, y=1

Explanation:

Replace y in the second equation with the right hand side of the irst equation to give $x + 7 = {(x + 3)}^{2} - 8$ $x+7=(x+3)^2-8$ . Expand the square on the right hand side $x + 7 = x^{2} + 6 x + 9 - 8$ $x+7=x^2+6x+9-8$ . Subtract $x + 7$ $x+7$ from both sides and tidy up $x^{2} + 5 x - 6 = 0$ $x^2+5x-6=0$ . Factorise $(x + 6) (x - 1) = 0$ $(x+6)(x-1)=0$ . This gives the two solutions $x = 1$ $x=1$ and $x = - 6$ $x=-6$ . Using the first equation, if $x = 1$ $x=1$ then $y = 8$ $y=8$ and if $x = - 6$ $x=-6$ then $y = 1$ $y=1$ .

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How do you solve the system $y = x + 7$ $y=x+7$ , $y = {(x + 3)}^{2} - 8$ $y=(x+3)^2-8$ ?

1 Answer

Explanation:

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How do you solve the system y=x+7y=x+7, y=(x+3)2−8 y=(x+3)^2-8?

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Explanation:

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How do you solve the system $y = x + 7$ $y=x+7$ , $y = {(x + 3)}^{2} - 8$ $y=(x+3)^2-8$ ?