##### Solving a system of linear and quadratic equations.

 Algebra Tutor: None Selected Time limit: 1 Day

Solve the following system of equations

y = x² + 7x - 5

y = 6x + 7

There may be more than one solution need all

Nov 14th, 2015

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$\\ y=x^2+7x-5---(1)\\ \\ y=6x+7---(2)$

Comparing equations (1) and (2) we get

$\\ x^2+7x-5=6x+7\\ \\ x^2+7x-5-6x-7=0\\ \\ x^2+x-12=0\\ \\ x^2+4x-3x-12=0\\ \\ x(x+4)-3(x+4)=0\\ \\ (x+4)(x-3)=0\\ \\ x+4=0\; \; \; or\; \; \; x-3=0\\ \\ x=-4\; \; \; or\; \; \; x=3$

Substitute x = -4 in equation (2)

$y=6\times(-4)+7=-24+7=-17$

Substitute x = 3 in equation (2)

$y=6\times3+7=18+7=25$

So the solution is (-4, -17) and (3, 25) i.e. the line intersects the parabola at 2 points.

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Nov 14th, 2015

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Nov 14th, 2015
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Nov 14th, 2015
Oct 24th, 2016
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