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8y=4x^(2)-12 x+46

y=(3)/(2)x+(5)/(4)
If 
(a,b) is the solution to the system of equations shown, what is the value of 
a ?

8y=4x212x+46 8 y=4 x^{2}-12 x+46 \newliney=32x+54 y=\frac{3}{2} x+\frac{5}{4} \newlineIf (a,b) (a, b) is the solution to the system of equations shown, what is the value of a a ?

Full solution

Q. 8y=4x212x+46 8 y=4 x^{2}-12 x+46 \newliney=32x+54 y=\frac{3}{2} x+\frac{5}{4} \newlineIf (a,b) (a, b) is the solution to the system of equations shown, what is the value of a a ?
  1. Substitute yy into first equation: We have a system of two equations:\newline11) 8y=4x212x+468y = 4x^2 - 12x + 46\newline22) y=32x+54y = \frac{3}{2}x + \frac{5}{4}\newlineTo find the solution to the system, we need to substitute the second equation into the first one to solve for xx.
  2. Distribute 88 on left side: Substitute yy from the second equation into the first equation:\newline8(32x+54)=4x212x+468\left(\frac{3}{2}x + \frac{5}{4}\right) = 4x^2 - 12x + 46
  3. Rearrange equation and solve: Distribute 88 on the left side of the equation:\newline8×(32)x+8×(54)=4x212x+468 \times \left(\frac{3}{2}\right)x + 8 \times \left(\frac{5}{4}\right) = 4x^2 - 12x + 46\newline12x+10=4x212x+4612x + 10 = 4x^2 - 12x + 46
  4. Divide and simplify: Rearrange the equation to set it to zero and solve for xx:4x212x12x10+46=04x^2 - 12x - 12x - 10 + 46 = 04x224x+36=04x^2 - 24x + 36 = 0
  5. Factor the quadratic equation: Divide the entire equation by 44 to simplify:\newlinex26x+9=0x^2 - 6x + 9 = 0
  6. Set factors equal to zero: Factor the quadratic equation: \newline(x - \(3)(x - 33) = 00
  7. Find xx value: Set each factor equal to zero and solve for xx:x3=0x - 3 = 0x=3x = 3
  8. Final solution for aa: We found the value of xx to be 33. Since we are looking for the value of aa, and (a,b)(a,b) is the solution to the system, a=xa = x.

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