Question

What values of \(y\) satisfy this system of equations? $$ \begin{array}{c} x=y^{2}-3 y+1 \\ 2 x=10 \end{array} $$ (A) \(-4\) and 2 (B) \(-1\) and 4 (C) 3 and 4 (D) 6 and 10

Short answer

The values of \(y\) that satisfy the system are \(-1\) and 4, so the answer is (B).

Step-by-step solution

Solve for x

The second equation of the system is given as \(2x = 10\). Divide both sides by 2 to solve for \(x\). This gives us \(x = 5\).

Substitute x into the first equation

Now that we know \(x = 5\), substitute this value into the first equation: \(x = y^2 - 3y + 1\). This becomes \(5 = y^2 - 3y + 1\).

Rearrange and simplify the equation

Rearrange the equation \(5 = y^2 - 3y + 1\) to standard quadratic form: \(y^2 - 3y + 1 - 5 = 0\). Simplifying gives \(y^2 - 3y - 4 = 0\).

Factor the quadratic equation

To factor \(y^2 - 3y - 4\), look for two numbers that multiply to -4 and add to -3. The numbers are -4 and 1. Thus, the equation factors to \((y - 4)(y + 1) = 0\).

Solve for y

Set each factor equal to zero to find the solutions for \(y\): \(y - 4 = 0\) gives \(y = 4\), and \(y + 1 = 0\) gives \(y = -1\).

Confirm the solution

The solutions for \(y\) are \(-1\) and \(4\), verifying that both numbers satisfy the initial system of equations. Check the available options to match the answer.

Key concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two in the form of \( ax^2 + bx + c = 0 \). They are solved to find the values of the variable that make the equation true. In our original problem, the quadratic equation was \( y^2 - 3y - 4 = 0 \). Here, the coefficients are \( a = 1 \), \( b = -3 \), and \( c = -4 \).

Quadratics can often be solved by several methods, such as:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

In this exercise, factoring was sufficient to find the solutions. Once in standard form, the equation was rearranged and simplified. The two roots or solutions of the quadratic equation turned out to be \( y = -1 \) and \( y = 4 \). These values satisfy the original system when put back into the equations.

System of Equations

A system of equations consists of multiple equations that share common variables. Solving a system requires finding values for the variables that satisfy all the equations simultaneously. In our exercise, the system consists of two equations:

• \( x = y^2 - 3y + 1 \)
• \( 2x = 10 \)

The goal was to find the value(s) of \( y \) that work in both equations. By solving for one variable and substituting it into the other equation, we can simplify and eventually solve the system.

In the context of our system, solving yielded \( x = 5 \) from the second equation. This value was substituted into the first equation to form a quadratic equation. Upon solving the resulting quadratic, we find the possible values for \( y \) that satisfy the original system.

Algebraic Factorization

Algebraic factorization is a method used to express a polynomial as a product of its factors. In our case, we factored the quadratic equation \( y^2 - 3y - 4 \) to solve for \( y \). Factorization often involves looking for two numbers that will multiply to give the constant term and add to give the middle coefficient.

For \( y^2 - 3y - 4 = 0 \), the two numbers are \(-4\) and \(1\) because:

• They multiply to \(-4 \times 1 = -4\)
• They add to \(-4 + 1 = -3\)

Thus, factorization gives \((y - 4)(y + 1) = 0\). Setting each factor equal to zero, \( y - 4 = 0 \) yields \( y = 4 \) and \( y + 1 = 0 \) yields \( y = -1 \). These are the solutions of the quadratic equation, hence solving the system of equations we started with.