SOLUTION: How do I solve this system of equations using substitution? {{{y=x^2-4x-1}}} {{{y=x+3}}}

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Question 646881: How do I solve this system of equations using substitution? y=x%5E2-4x-1 y=x%2B3

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y=x%5E2-4x-1............1
y=x%2B3.................2
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y=x%2B3.................2...here you have y on one side, substitute it in 1

x%2B3=x%5E2-4x-1............1...solve for x
1%2B3=x%5E2-4x-x
4=x%5E2-5x
0=x%5E2-5x-4
or x%5E2-5x-4=0.......use quadratic formula to find x
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%2A1%2A%28-4%29+%29%29%2F%282%2A1%29+
x+=+%285+%2B-+sqrt%2825%2B16+%29%29%2F2+
x+=+%285+%2B-+sqrt%2841%29%29%2F2+
x+=+%285+%2B-+6.4%29%2F2+
solutions:
x+=+%285+%2B+6.4%29%2F2+
x+=+11.4%2F2+
x+=+5.7+
or
x+=+%285+-+6.4%29%2F2+
x+=+-1.4%2F2+
x+=+-0.7+
now find y

y=x%2B3.................2...if x+=+5.7+
y=5.7%2B3

y=8.7
y=x%2B3.................2...if x+=+-0.7+
y=-0.7%2B3

y=2.3

your pairs of solutions are:
x+=+5.7+ , y=8.7
and
x+=+-0.7+, y=2.3