$\begin{align} y &= 6x-30 \\ y &= x^{2}-18x+114 \end{align}$ $\newline$ If $(a,b)$ is the solution to the system of equations shown, what is the value of $b$ ? $\newline$ $\square$

View step-by-step help

Home

Math Problems

Algebra 2

Evaluate expression when a complex numbers and a variable term is given

Full solution

\begin{align*} y &= 6x-30 \\ y &= x^{2}-18x+114 \end{align*}

\newline

(a,b)

is the solution to the system of equations shown, what is the value of

b

\newline

\square

Set Equations Equal: Given the system of equations: $\newline$ $1$ ) $y = 6x - 30$ $\newline$ $2$ ) $y = x^2 - 18x + 114$ $\newline$ To find the value of $b$ , we need to solve the system of equations. We can do this by setting the two expressions for $y$ equal to each other and solving for $x$ .
Move Terms to Solve: Set the two equations equal to each other: $\newline$ $6x - 30 = x^2 - 18x + 114$ $\newline$ Now, we will move all terms to one side to solve for $x$ .
Rearrange and Combine: Rearrange the equation: $\newline$ $x^2 - 18x + 114 - 6x + 30 = 0$ $\newline$ Combine like terms: $\newline$ $x^2 - 24x + 144 = 0$ $\newline$ This is a quadratic equation in standard form.
Factor Quadratic Equation: Factor the quadratic equation: $\newline$ $(x - 12)(x - 12) = 0$ $\newline$ This gives us a repeated root, meaning $x$ has one value where the equation is true.
Solve for x: Solve for x: $\newline$ $x - 12 = 0$ $\newline$ $x = 12$ $\newline$ We have found the value of $x$ that satisfies both equations.
Substitute $x$ into Equation: Now that we have the value of $x$ , we can substitute it back into either of the original equations to find the value of $y$ , which corresponds to $b$ in the solution $(a,b)$ . Let's substitute $x$ into the first equation: $y = 6x - 30$
Calculate Value of y: Substitute $x = 12$ into the equation: $\newline$ $y = 6(12) - 30$ $\newline$ Calculate the value of y: $\newline$ $y = 72 - 30$ $\newline$ $y = 42$ $\newline$ We have found the value of y, which is the value of $b$ in the solution $(a,b)$ .