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Solve the simultaneous equations: \[ \begin{array}{l} 2 x+3 y=2 \\ 3 x^{2}+4 x ...
Jul 31, 2024
Solution by Steps
step 1
We start with the given simultaneous equations: 2x+3y=2 2x + 3y = 2 3x2+4xy−y2−12=0 3x^2 + 4xy - y^2 - 12 = 0
step 2
Solve the first equation for y y : 2x+3y=2 2x + 3y = 2 3y=2−2x 3y = 2 - 2x y=2−2x3 y = \frac{2 - 2x}{3}
step 3
Substitute y=2−2x3 y = \frac{2 - 2x}{3} into the second equation: 3x2+4x(2−2x3)−(2−2x3)2−12=0 3x^2 + 4x \left( \frac{2 - 2x}{3} \right) - \left( \frac{2 - 2x}{3} \right)^2 - 12 = 0 3x2+8x−8x23−(2−2x)29−12=0 3x^2 + \frac{8x - 8x^2}{3} - \frac{(2 - 2x)^2}{9} - 12 = 0
step 4
Simplify the equation: 3x2+8x−8x23−4−8x+4x29−12=0 3x^2 + \frac{8x - 8x^2}{3} - \frac{4 - 8x + 4x^2}{9} - 12 = 0 3x2+8x−8x23−4−8x+4x29−12=0 3x^2 + \frac{8x - 8x^2}{3} - \frac{4 - 8x + 4x^2}{9} - 12 = 0 3x2+24x−24x29−4−8x+4x29−12=0 3x^2 + \frac{24x - 24x^2}{9} - \frac{4 - 8x + 4x^2}{9} - 12 = 0 3x2+24x−24x2−4+8x−4x29−12=0 3x^2 + \frac{24x - 24x^2 - 4 + 8x - 4x^2}{9} - 12 = 0 3x2+32x−28x2−49−12=0 3x^2 + \frac{32x - 28x^2 - 4}{9} - 12 = 0
step 5
Combine like terms and solve for x x : 3x2+32x−28x2−49−12=0 3x^2 + \frac{32x - 28x^2 - 4}{9} - 12 = 0 27x2+32x−28x2−4−108=0 27x^2 + 32x - 28x^2 - 4 - 108 = 0 −x2+32x−112=0 -x^2 + 32x - 112 = 0 x2−32x+112=0 x^2 - 32x + 112 = 0 (x−4)(x−28)=0 (x - 4)(x - 28) = 0 x=4orx=28 x = 4 \quad \text{or} \quad x = 28
step 6
Substitute x=4 x = 4 and x=28 x = 28 back into y=2−2x3 y = \frac{2 - 2x}{3} to find y y : For x=4 x = 4 : y=2−2(4)3=2−83=−63=−2 y = \frac{2 - 2(4)}{3} = \frac{2 - 8}{3} = \frac{-6}{3} = -2 For x=28 x = 28 : y=2−2(28)3=2−563=−543=−18 y = \frac{2 - 2(28)}{3} = \frac{2 - 56}{3} = \frac{-54}{3} = -18
Answer
The solutions are (x,y)=(4,−2) (x, y) = (4, -2) and (28,−18) (28, -18)
Key Concept
Solving simultaneous equations
Explanation
To solve simultaneous equations, we can use substitution or elimination methods. Here, we used substitution to find the values of x x and y y .
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