Problem 84 - College Algebra with Modeling and Visualization

Question

Solve each system. $$ \begin{aligned} -2 y^{2} &=x-5 \\ y^{2} &=2 x \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \(x = 1\), \(y = \pm \sqrt{2}\).

Step 1: Set the equations

We are given two equations in terms of two variables, \(x\) and \(y^2\):1. \(-2y^2 = x - 5\)2. \(y^2 = 2x\).Our goal is to find the values of \(x\) and \(y\) that satisfy both equations.

Step 2: Substitute one equation into the other

Substitute \(y^2 = 2x\) from the second equation into the first equation. This will give us a system in terms of one variable:Substitute \(y^2 = 2x\) into \(-2y^2 = x - 5\):\[-2(2x) = x - 5\].

Step 3: Simplify and solve for x

Simplify the equation from the substitution:\[-4x = x - 5\].Add \(4x\) to both sides of the equation:\[0 = 5x - 5\].Add 5 to both sides:\[5 = 5x\].Divide both sides by 5 to solve for \(x\):\[x = 1\].

Step 4: Use the value of x to find y^2

Substitute \(x = 1\) back into the equation \(y^2 = 2x\) to solve for \(y^2\):\[y^2 = 2(1) = 2\].

Step 5: Solve for y

Since \(y^2 = 2\), take the square root of both sides to find \(y\):\[y = \pm \sqrt{2}\].

Step 6: Verify the solution

Plug \(x = 1\) and \(y = \pm \sqrt{2}\) back into both original equations to verify that they satisfy both equations.1. For \(-2y^2 = x - 5\): - When \(y^2 = 2\), \(-2(2) = 1 - 5\) simplifies to \(-4 = -4\).2. For \(y^2 = 2x\): - \(2 = 2(1)\) simplifies to \(2 = 2\).Both sets of values satisfy the original system.

Key Concepts

Substitution Method

The substitution method is a strategy for solving systems of equations where we solve one equation for one variable and then substitute that expression into the other equation. This helps us reduce the equations to a single variable, making it easier to solve.

In our exercise, we were given the equations:

• \(-2y^2 = x - 5\)
• \(y^2 = 2x\)

We began by isolating one of the variables. From the second equation, \(y^2 = 2x\), we solved for \(y^2\). This expression was then substituted into the first equation. This substitution allowed us to transform it to a single-variable equation:\[-2(2x) = x - 5\].

The beauty of the substitution method lies in its straightforward approach. By eliminating one variable, we simplify the math and make it easier to find the value of the other variable, narrowing down our search for the solution.

Algebraic Manipulation

Algebraic manipulation helps us simplify equations and solve them systematically. This involves using basic algebraic rules such as addition, subtraction, multiplication, and division to rearrange and simplify expressions.

From our substituted equation \[-4x = x - 5\],we used algebraic manipulation to isolate the variable \(x\).

• Add \(4x\) to both sides to group \(x\) terms: \[0 = 5x - 5\].
• Add \(5\) to both sides to eliminate constant terms on one side:\[5 = 5x\].
• Divide by \(5\) to solve for \(x\):\[x = 1\].

These steps highlight how algebraic manipulation is crucial for simplifying and solving equations. Each operation helps to gradually isolate the variable, leading us directly to the solution of the equation.

Without these handy operations, finding precise solutions would be more complex and cumbersome.

Square Root Method

The square root method is an essential tool when dealing with quadratic expressions, particularly when they appear as squared terms. After determining the value of \(x\), we return to our expression involving \(y^2\).

Substituting \(x = 1\) into the equation \(y^2 = 2x\), we find:\[y^2 = 2(1) = 2\].
To extract the value for \(y\), we take the square root of both sides:\[y = \pm \sqrt{2}\].

This step is key because each squared term has two possible roots—positive and negative. Thus, \(y\) equals both \(\sqrt{2}\) and \(-\sqrt{2}\).

• This method allows us to consider all possible solutions for \(y\), considering its quadratic nature.
• Both solutions must often be validated back with the original equations to ensure they are viable.

Using the square root method, we gain a more complete and accurate understanding of solutions involving squared terms.