Problem 51 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}-y=0 \\ x^{2}-4 x+y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((0, 0)\) and \((2, 4)\).

Step 1: Express y from first equation

Start with the first equation: \(x^2 - y = 0\). Rearrange this equation to solve for \(y\): \(y = x^2\). Save this expression for use in the second equation.

Step 2: Substitute y in the second equation

Substitute \(y = x^2\) from Step 1 into the second equation \(x^2 - 4x + y = 0\). The equation becomes: \(x^2 - 4x + x^2 = 0\).

Step 3: Simplify the substituted equation

Combine like terms in the substituted equation: \(x^2 + x^2 - 4x = 0\) simplifies to \(2x^2 - 4x = 0\).

Step 4: Factorize the equation

Factorize the equation \(2x^2 - 4x = 0\) by taking out the common factor of \(2x\): \(2x(x - 2) = 0\).

Step 5: Solve the factored equation

Set each factor to zero: \(2x = 0\) gives \(x = 0\) and \(x - 2 = 0\) gives \(x = 2\).

Step 6: Solve for y values

Substitute \(x = 0\) and \(x = 2\) back into the expression \(y = x^2\) from Step 1:- For \(x = 0\), \(y = 0^2 = 0\).- For \(x = 2\), \(y = 2^2 = 4\).

Step 7: Write the solutions as pairs

The solutions to the system of equations are \((x, y) = (0, 0)\) and \((2, 4)\).

Key Concepts

Solving systems of equations

When faced with a system of equations, the goal is to find values of the variables that make all the equations true at the same time. Each equation in the system represents a relationship between the variables, and solving the system involves finding a common solution. In this exercise, we have two equations involving variables \(x\) and \(y\):

• \(x^2 - y = 0\)
• \(x^2 - 4x + y = 0\)

To find solutions for both \(x\) and \(y\), we need to work with both equations. Solving systems can be approached with different methods, but the substitution method is often particularly straightforward for systems involving quadratic equations. Systems of equations can have one solution, many solutions, or no solution at all, depending on how the equations relate to each other.

Factoring quadratic equations

Factoring is a powerful technique used to solve quadratic equations. In our problem, we reach a quadratic equation after substituting \(y = x^2\) into the second equation:\[2x^2 - 4x = 0\]. Quadratic equations are typically in the form \(ax^2 + bx + c = 0\), and they can often be simplified by factoring, which involves writing the equation as a product of its linear factors.
To factor \(2x^2 - 4x = 0\), we first look for the greatest common factor among all the terms. Here, the common factor is \(2x\), allowing us to rewrite the equation as:\[2x(x - 2) = 0\]. This factored form reveals the possible solutions directly, indicating where the expression equals zero. After factoring, each factor can be set individually to zero to find the specific \(x\) values.

Substitution method

The substitution method is especially useful for solving systems when one of the equations can be easily solved for one variable. Once an equation is solved for one variable, this expression is substituted into the other equation. This method was applied in our exercise by solving the first equation \(x^2 - y = 0\) for \(y\), which yields \(y = x^2\).
After finding \(y\) in terms of \(x\), we substituted \(y = x^2\) into the second equation: \(x^2 - 4x + y = 0\). This gave us a simpler equation in terms of only \(x\): \(x^2 - 4x + x^2 = 0\) which simplified to \(2x^2 - 4x = 0\). By reducing the system to a single variable, solving becomes more manageable. Once \(x\) values were found, they were used to calculate corresponding \(y\) values by substituting back into the expression for \(y\). This method shows how substitution can simplify solving systems of equations, making it a handy tool when equations involve multiple variables.