Problem 28 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} x^{2}+y^{2}=6 \\ 5 x^{2}+y^{2}=10 \end{array}$$

Step-by-Step Solution

Answer

The solutions to the given system of equations are \((1, \sqrt{5})\), \((1, -\sqrt{5})\), \((-1, \sqrt{5})\), and \((-1, -\sqrt{5})\).

Step 1: Solve the first equation for y.

Rearrange the first equation to make y the subject: \[y^2 = 6 - x^2.\] So, we have found an expression for \(y^2\).

Step 2: Substitute the expression for \(y^2\) into the second equation.

Replace \(y^2\) in the second equation with the expression we found in Step 1: \[5x^2 + (6 - x^2) = 10.\]

Step 3: Simplify and solve for x.

Combine the terms and solve for x: \[4x^2 = 4\Rightarrow x^2 = 1.\] Taking the square root of both sides, we find that \(x = \pm 1\).

Step 4: Substitute the obtained x values back into the expression for y.

For each value of x, substitute it into the expression for \(y^2\) we got in Step 1: - If \(x = 1\), then \(y^2 = 6 - 1^2 = 5\) ⇒ \(y = \pm\sqrt{5}\). - If \(x = -1\), then \(y^2 = 6 - (-1)^2 = 5\) ⇒ \(y = \pm\sqrt{5}\).

Step 5: Write the solutions.

We now have four possible combinations for the intersection points: \[(1, \sqrt{5}),\quad (1, -\sqrt{5}),\quad (-1, \sqrt{5}),\quad (-1, -\sqrt{5}).\] So, the solutions to the given system of equations are \((1, \sqrt{5})\), \((1, -\sqrt{5})\), \((-1, \sqrt{5})\), and \((-1, -\sqrt{5})\).

Key Concepts

Understanding Quadratic Equations

Quadratic equations are equations where the highest power of the variable is two. In general, a quadratic equation in one variable is of the form:

• \( ax^2 + bx + c = 0 \)

where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). A system of quadratic equations involves having more than one equation with quadratic terms that must be solved at the same time. In the context of our exercise, we were given two equations, both featuring terms with \(x^2\) and \(y^2\).
This introduces two variables which can be a bit more complex to solve because the relationship between \(x\) and \(y\) is not linear, thus often resulting in multiple solutions.
By carefully manipulating these equations, we can find pairs of \((x, y)\) values that satisfy both equations.

The Role of Algebraic Manipulation

Algebraic manipulation involves using mathematical operations to rearrange, simplify, or solve equations. This is a crucial skill, as it allows you to isolate variables, simplify expressions, and ultimately find solutions to equations.
In our problem, algebraic manipulation was used to express one variable in terms of the other. In step 1, the equation \(x^2 + y^2 = 6\) was rearranged to solve for \(y^2\), giving us \(y^2 = 6 - x^2\).
This expression was crucial in our ability to substitute and solve further, as it simplified the system of equations from two variables to one. Key manipulations like these focus on rearranging equations and keeping operations balanced on both sides to maintain equality.

• Substitution is streamlined with clean expressional rearrangements.
• Isolating terms helps in identifying and extracting possible solutions sequentially.

These strategies can be applied to numerous algebra problems, assisting in making complex equations more manageable.

Solving Systems by Substitution

Solving systems of equations by substitution is an effective method when one equation can be easily solved for one variable. Once you express one variable in terms of another, you can substitute this expression into other equations.
This is precisely what we did in this exercise. After identifying that \(y^2 = 6 - x^2\) in the original first equation, this expression was substituted into the second equation, resulting in a single variable equation:

• \(5x^2 + (6 - x^2) = 10\)

With only the variable \(x\) to consider, it became substantially easier to find \(x\). Solving this gave us \(x^2 = 1\), leading to \(x = \pm 1\).
Substituting these values back into the expression for \(y^2\) gave us possible \(y\)-values, creating the solution pairs. The substitution method is valuable because it reduces the complexity of solving multiple variables at once by breaking the system down into simpler, successive problems.