How do you solve the system $Y = x + 7$ $Y=x+7$ , $x^{2} + y^{2} = 25$ $x^2+y^2=25$ ?

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How do you solve the system $Y = x + 7$ $Y=x+7$ , $x^{2} + y^{2} = 25$ $x^2+y^2=25$ ?

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Answer 1 · 2016-01-01T17:21:29

Substitute $y = x + 7$ $y = x+7$ into $x^{2} + y^{2} = 25$ $x^2+y^2 = 25$ to get a quadratic in $x$ $x$ , hence solutions:

$(x, y) = (- 3, 4) or (- 4, 3)$ $(x, y) = (-3, 4) or (-4, 3)$

Explanation:

Substitute $y = x + 7$ $y = x+7$ into $x^{2} + y^{2} = 25$ $x^2+y^2 = 25$ ...

$25 = x^{2} + y^{2}$ $25= x^2+y^2$

$= x^{2} + {(x + 7)}^{2}$ $=x^2+(x+7)^2$

$= x^{2} + x^{2} + 14 x + 49$ $=x^2+x^2+14x+49$

$= 2 x^{2} + 14 x + 49$ $=2x^2+14x+49$

Subtract $25$ $25$ from both ends to get:

$2 x^{2} + 14 x + 24 = 0$ $2x^2+14x+24 = 0$

Divide through by $2$ $2$ to get:

$x^{2} + 7 x + 12 = 0$ $x^2+7x+12 = 0$

We can factor this by finding a pair of factors of $12$ $12$ whose sum is $7$ $7$ . The pair $3, 4$ $3, 4$ works.

Hence:

$0 = x^{2} + 7 x + 12 = (x + 3) (x + 4)$ $0 = x^2+7x+12 = (x+3)(x+4)$

This has roots $x = - 3$ $x=-3$ and $x = - 4$ $x=-4$ , corresponding to $y = 4$ $y = 4$ and $y = 3$ $y=3$ respectively.

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How do you solve the system $Y = x + 7$ $Y=x+7$ , $x^{2} + y^{2} = 25$ $x^2+y^2=25$ ?

1 Answer

Explanation:

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How do you solve the system $Y = x + 7$ $Y=x+7$ , $x^{2} + y^{2} = 25$ $x^2+y^2=25$ ?