Question

If \(2x^2 + 10 = 7y - 5x\) and \(4y^2 - 5y = -(7x + 17)\), then find the value of \((x + y)\)?

Hint:

When a system of equations looks complex, try adding or subtracting them. This can sometimes eliminate terms or reveal a simpler structure, like a sum of squares equalling zero, which provides a unique solution.

Answer 1

Step 1: Understanding the Concept:
We are given a system of two non-linear equations. A common strategy for such systems is to manipulate and combine them in a way that leads to a simpler equation. Here, adding the two equations reveals a hidden structure.
Step 2: Detailed Explanation:
First, let's rearrange both equations to the standard form \((\dots) = 0\).
Equation 1: \[ 2x^2 + 10 = 7y - 5x \implies 2x^2 + 5x - 7y + 10 = 0 \] Equation 2: \[ 4y^2 - 5y = -(7x + 17) \implies 4y^2 - 5y = -7x - 17 \implies 4y^2 - 5y + 7x + 17 = 0 \] Now, add the two rearranged equations together: \[ (2x^2 + 5x - 7y + 10) + (4y^2 - 5y + 7x + 17) = 0 \] Group the like terms: \[ 2x^2 + 4y^2 + (5x + 7x) + (-7y - 5y) + (10 + 17) = 0 \] \[ 2x^2 + 4y^2 + 12x - 12y + 27 = 0 \] This equation can be simplified by completing the square for both \(x\) and \(y\) terms. Group the \(x\) and \(y\) terms: \[ (2x^2 + 12x) + (4y^2 - 12y) + 27 = 0 \] Factor out the coefficients of the squared terms: \[ 2(x^2 + 6x) + 4(y^2 - 3y) + 27 = 0 \] Complete the square inside the parentheses. For \(x^2+6x\), we add and subtract \((\frac{6}{2})^2 = 9\). For \(y^2-3y\), we add and subtract \((\frac{-3}{2})^2 = \frac{9}{4}\). \[ 2(x^2 + 6x + 9 - 9) + 4(y^2 - 3y + \frac{9}{4} - \frac{9}{4}) + 27 = 0 \] \[ 2((x+3)^2 - 9) + 4((y-\frac{3}{2})^2 - \frac{9}{4}) + 27 = 0 \] Distribute the coefficients: \[ 2(x+3)^2 - 18 + 4(y-\frac{3}{2})^2 - 9 + 27 = 0 \] \[ 2(x+3)^2 + 4(y-\frac{3}{2})^2 + (-18 - 9 + 27) = 0 \] \[ 2(x+3)^2 + 4(y-\frac{3}{2})^2 = 0 \] Since we are dealing with real numbers, the terms \((x+3)^2\) and \((y-\frac{3}{2})^2\) are always non-negative. The sum of non-negative terms can be zero only if each term is individually zero. \[ (x+3)^2 = 0 \implies x + 3 = 0 \implies x = -3 \] \[ (y-\frac{3}{2})^2 = 0 \implies y - \frac{3}{2} = 0 \implies y = \frac{3}{2} \] Step 3: Final Answer:
We are asked to find the value of \((x+y)\). \[ x + y = -3 + \frac{3}{2} = -\frac{6}{2} + \frac{3}{2} = -\frac{3}{2} \]