Question 224914
solve this simultaneous equation: 

{{{x+y=10}}}  Equation 1
{{{y=3x^2-2x-3}}}  Equation 2


Step 1.  In Equation 1, solve for {{{y=10-x}}} and substitute y into Equation 2.


{{{10-x=3x^2-2x-3}}}


Step 2.  Add x-10 to both sides of the equatiion


{{{10-x+x-10=3x^2-2x-3+x-10}}}


{{{0=3x^2-x-13}}}


Step 3.  To solve use the quadratic formula given below:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


where a=3, b=-1, and c=-13


*[invoke quadratic "x", 3, -1, -13 ]


Step 4.  Now, for each value of x found in Step 3, substitute into y=10-x to get two values for y.


I hope the above steps were helpful. 


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And good luck in your studies!


Respectfully,
Dr J


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