How do you solve this system of equations: $y= 14x ^ { 2} + 12x - 28 and y = 12x - 14$ ?

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Answer 1 · 2017-12-04T01:42:17

$(-1,-26)$ and $(1,-2)$

$y=14x^2+12x-28$ and $y=12x-14$

$14x^2+12x-28=12x-14$

$14x^2+12x-12x-28+14$

$14x^2-14=0$

$14(x^2-1)=0$

$(x^2-1)=0$

$x=+-1$

for x=1

$y=14(1)^2+12(1)-28$

$y=14+12-28 -> y=-2$ $(1,-2)$

for x=-1

$y=12(-1)-14=-12-14=-26 -> y=-26$

$(-1,-26)$ and $(1,-2)$

How do you solve this system of equations: y= 14x ^ { 2} + 12x - 28 and y = 12x - 14y= 14x ^ { 2} + 12x - 28 and y = 12x - 14?