Question 224915
Solve this simultaneous equation:

{{{y=5-4x^2}}}  Equation 1
{{{y+3x=6}}}  or {{{y=6-3x}}}  Equation 2


Step 1.  Set Equations 1 and 2 to be equal.


{{{5-4x^2=6-3x}}}


Step 2.  Add {{{4x^2-5}}} to both sides of the equation


{{{5-4x^2+4x^2-5=6-3x+4x^2-5}}}


{{{0=4x^2-3x+1}}}


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


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


where a=4, b=-3, and c=1


*[invoke quadratic "x", 4, -3, 1 ]


Step 4.  For each value of x found in Step 3, substitute each value into y=6-3x to get a total of 2 values for y.  You now have two points as your solution to these equations.  Your solutions will be complex numbers.


I hope the above steps were helpful. 


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And good luck in your studies!


Respectfully,
Dr J


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