Problem 22 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &x^{2}+y^{2}=49\\\ &x-2 y^{2}=7 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations \(\begin{aligned} &x^{2}+y^{2}=49\\ &x-2 y^{2}=7 \end{aligned}\) is the ordered pair \((7, 0)\).

Step 1: Isolate a variable from either equation

We can isolate x from the second equation to get x in terms of y: \(x = 2y^2 + 7\)

Step 2: Substitute the expression obtained in Step 1 into the first equation

Replace x in the first equation (circle) with the expression found in Step 1: \((2y^2 + 7)^2 + y^2 = 49\)

Step 3: Simplify and solve the second equation for y

Expand and simplify the equation: \(4y^4 + 28y^2 + 49 + y^2 = 49\) Combine like terms: \(4y^4 + 29y^2 = 0\) Factor out the common factor of y^2: \(y^2(4y^2 + 29) = 0\) Solve for y: \(y^2 = 0\) or \(4y^2 + 29 = 0\) We find that \(y = 0\) or \(y^2 = -\dfrac{29}{4}\). Since y^2 cannot be negative, we discard the second root and determine that the only solution for y is \(y = 0\).

Step 4: Find the value of x using the expression obtained in Step 1

Substitute y = 0 into the expression for x: \(x = 2(0)^2 + 7\) Therefore, \(x = 7\).

Step 5: Write the solution as an ordered pair

Since we found that x = 7 and y = 0, the solution to this system of equations is the ordered pair \((7, 0)\).