I need help on Algebra 2, "Solving Nonlinear Systems" substitution/elimination?
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Solve each system by using the substitution method. 2x^2 - y^2 = 14 y - 2x = -4 i looked in the back of my Algebra 2 book and the answer is x = 3,5 y = 2,6 when i tried to solve it, i ended up with -2x^2 - 16x + 2 = 0, i dont know what to do after that :\
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Answer:
2x^2 - y^2 = 14 y - 2x = -4 Solve the second equation for y by adding 2x to both sides: y = 2x - 4 Plug that into the first equation and solve for x: 2x^2 - (2x - 4)^2 = 14 2x^2 - (4x^2 - 16x + 16) = 14 2x^2 - 4x^2 + 16x - 16 = 14 -2x^2 + 16x - 30 = 0 If you divide both sides by -2, x^2 - 8x + 15 = 0 What multiplies to 1*15 and adds to -8? -3 and -5 work: (x - 3)(x - 5) = 0 x = 3, 5 Plug each of those into the second equation and find the corresponding y-values (you could plug them into the first, but that seems harder): y = 2x - 4 When x = 3, y = 2*3 - 4 = 6 - 4 = 2 When x = 5, y = 2*5 - 4 = 10 - 4 = 6
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Other answers
Substitution will be best here; however, I prefer elimination hands out. y = 2x - 4 2x^2 - (2x-4)(2x-4) = 14 2x^2 - [ 4x^2 - 16x + 16 ] = 14 2x^2 - 4x^2 + 16x - 16 = 14 - 2x^2 + 16x - 30 = 0 -2(x^2 - 8x + 15) LCD: 15 Common Multiples: - 5, - 3 ( x - 5 ) ( x - 3 ) x = 5, 3 You can go ahead and solve for Y here.
Andrew
y=2x-4 (1) y^2=2x^2-14 (2) From (1) y^2=4x^2-16x+16 So 4x^2-16x+16=2x^2-14 2x^2-16x+30=0 x^2-8x+15=0 Factorise x^2-3x-5x+15=0 x(x-3)-5(x-3)=0 (x-5)(x-3)=0 x=5 or 3 Substitute into y-2x=-4 y=2x-4 y=6 or 2 Hope this helps
Jack
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