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Difficulty: MediumNonlinear Systems of Equations

If (x,y)(x, y) is a solution to the system of equations below and x>1x > 1, what is the value of xx?

y−2x=3y2−5x2=20\begin{aligned} y - 2x &= 3 \\ y^2 - 5x^2 &= 20 \end{aligned}

Answer: 11

Answer

The value of xx is 11.
The system can be solved by substituting the first equation into the second. Rearranging the first equation gives y=2x+3y = 2x + 3. Substituting this expression for yy in the second equation gives (2x+3)2−5x2=20(2x + 3)^2 - 5x^2 = 20. Expanding the binomial yields 4x2+12x+9−5x2=204x^2 + 12x + 9 - 5x^2 = 20, which simplifies to −x2+12x+9=20-x^2 + 12x + 9 = 20. Rearranging into standard quadratic form gives x2−12x+11=0x^2 - 12x + 11 = 0. Factoring the quadratic equation results in (x−11)(x−1)=0(x - 11)(x - 1) = 0. Thus, x=11x = 11 or x=1x = 1. Since the problem specifies x>1x > 1, the value of xx must be 11.

Step-by-Step Solution

1
Rearrange the first equation to express yy in terms of xx.
y=2x+3y = 2x + 3
This allows for substitution into the second equation.
2
Substitute y=2x+3y = 2x + 3 into the second equation.
(2x+3)2−5x2=20(2x + 3)^2 - 5x^2 = 20
To eliminate yy and obtain an equation in terms of xx only.
3
Expand and simplify the equation.
x2−12x+11=0x^2 - 12x + 11 = 0
Expanding (2x+3)2(2x + 3)^2 gives 4x2+12x+94x^2 + 12x + 9. Subtracting 5x25x^2 yields −x2+12x+9=20-x^2 + 12x + 9 = 20. Subtracting 20 from both sides gives −x2+12x−11=0-x^2 + 12x - 11 = 0. Multiplying the entire equation by −1-1 results in standard form.
4
Factor the quadratic equation.
(x−11)(x−1)=0(x - 11)(x - 1) = 0
To find the possible values of xx.
5
Apply the given constraint x>1x > 1.
x=11x = 11
The solutions are x=11x = 11 and x=1x = 1. The constraint x>1x > 1 excludes x=1x = 1, leaving x=11x = 11 as the only valid solution.

Key Concept

Solving a system of nonlinear equations by substitution and factoring a quadratic equation.
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Topics

Nonlinear Systems of Equations

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