Problem 23 - 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 \((7, 0)\).

Key Concepts

Simultaneous Equations

Simultaneous equations, also known as systems of equations, involve finding values for variables that satisfy multiple equations at the same time. In this case, two equations describe a relationship between the variables \(x\) and \(y\). Solving these equations involves finding a point or points that lie on both equations. This means we are looking for values of \(x\) and \(y\) that make both equations true simultaneously.
In the exercise provided, one equation describes a circle (since it has the form \(x^2 + y^2 = r^2\)), and the other appears to be a linear equation in terms of \(x\). The task is to find common solutions, or intersections, of these two equations. This is achieved by finding the values that work for both equations simultaneously.

Substitution Method

The substitution method is a straightforward technique for solving simultaneous equations. It involves isolating one variable in one equation and then substituting that expression into the other equation. This simplifies the problem to a single equation with one variable.
In our example, we started with the second equation, \(x - 2y^2 = 7\). We isolated \(x\), getting \(x = 2y^2 + 7\). We then substituted this expression into the first equation \(x^2 + y^2 = 49\). By replacing \(x\) with \(2y^2 + 7\), we can focus on solving for the variable \(y\) first. This technique simplifies the complex problem of dealing with two equations and two variables.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate and solve for a variable. This concept is critical in solving systems of equations using the substitution method.
After substituting for \(x\) in the first equation, we have an expression \((2y^2 + 7)^2 + y^2 = 49\). This requires expanding the square and combining like terms to obtain \(4y^4 + 29y^2 = 0\). Factoring \(y^2\) from the equation simplifies it further to \(y^2(4y^2 + 29) = 0\).
Through algebraic manipulation, we solve that \(y^2 = 0\), giving us \(y = 0\) as a possible real solution. The other part of the factored equation, \(4y^2 + 29 = 0\), gives no real solution because it would imply \(y^2\) is negative, which isn't possible for real numbers.

Ordered Pairs Solution

An ordered pairs solution consists of the values of \(x\) and \(y\) that satisfy both equations in a system. This pair represents a point of intersection on the graphs of the equations.
With \(y = 0\) found through manipulation, we substitute back into the isolated expression for \(x\) from earlier, \(x = 2y^2 + 7\). Since \(y = 0\), substituting gives \(x = 7\). This gives the ordered pair solution \((x, y) = (7, 0)\).
This ordered pair tells us that the point \((7, 0)\) is where both equations intersect. Therefore, this solution is the set of values for \(x\) and \(y\) that satisfy both the original equations we started with.