# 9.2: X the Unknown

Difficulty Level: At Grade Created by: CK-12

\begin{align*}X\end{align*} The Unknown – Solve for Unknowns using the Distributive Property

Teacher Notes

Students apply the distributive property and the rule for order of operations to solve equations involving unknowns. In all problems, students have to find square roots of values. All roots are the positive roots of square numbers. Be sure that students show all computational steps.

Solutions:

\begin{align*}\textbf{\textit{X}}\end{align*} The Unknown 1

\begin{align*}x^2 + 4(3 + x) - 3 = 2(2x + 9)\\ x^2 + 12 + 4x - 3 = 4x + 18\\ x^2 + 12 - 3 - 4x = 4x - 4x + 18\\ x^2 + 12 - 3 = 18\\ x^2 + 9 = 18\\ x^2 + 9 - 9 = 18 - 9\\ x^2 = 9\\ x=3\end{align*}

\begin{align*}\textbf{\textit{X}}\end{align*} The Unknown 2

\begin{align*}4(3 - x) + 9^2 = x(x - 4) + 7^2 - 5\\ 12 - 4x + 9^2 = x^2 - 4x + 7^2 - 5\\ 12 - 4x + 81 = x^2 - 4x + 49 - 5\\ 93 - 4x = x^2 - 4x + 44\\ 93 - 4x + 4x = x^2 - 4x + 44 + 4\\ 93 = x^2 + 44\\ 93 - 44 = x^2 + 44 - 44\\ 49 = x^2\\ 7 = x\end{align*}

\begin{align*}\textbf{\textit{X}}\end{align*} The Unknown 3

\begin{align*}5(x + 8) + 8^2 - (5^2 + 3 \times 13) = x(x+5) + (9-7)^2\\ 5x + 40 + 8^2 - (5^2 + 3 \times 13) = x^2 + 5x + (9-7)^2\\ 5x + 40 + 8^2 - (25 + 3 \times 13) = x^2 + 5x + 2^2\\ 5x + 40 + 64 - (25 + 39) = x^2 + 5x + 4\\ 5x + 40 + 64 - 64 = x^2 + 5x + 4\\ 5x + 40 = x^2 + 5x + 4\\ 5x + 40 - 5x = x^2 + 5x + 4 - 5x\\ 40 = x^2 + 4\\ 40 - 4 = x^2 + 4 - 4\\ 36 = x^2\\ 6 = x\end{align*}

\begin{align*}2(x + 8) + 2x = x(x + 2^2) + 3^2 - 9\end{align*}

What is the value of \begin{align*}x\end{align*}?

\begin{align*}&\mathbf{Describe:} && \text{The equation has one unknown}, \ x. \ \text{There are two sets of parentheses.}\\ \\ &\mathbf{My \ Job:} && \text{Do the computations to figure out the value of} \ x.\\ \\ &\mathbf{Plan :} && \text{Follow the order of operations to simplify the equation. Then solve for the value}\\ &&& \text{of} \ x. \ \text{Order of operations: Parentheses first. Apply the distributive property.}\\ &&& \text{Then do all computations inside of the parentheses next, following the rule for the}\\ &&& \text{order of operations.}\\ &&& \qquad \text{Parentheses}\\ &&& \qquad \text{Exponents}\\ &&& \qquad \text{Multiplication and Division, left to right}\\ &&& \qquad \text{Addition and Subtraction, left to right}\\ \\ &\mathbf{Solve:} && \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2( x + 8) + 2x = x(x + 2^2) + 3^2 - 9\\ &&& \mathbf{Distributive \ Property} \qquad \qquad \quad 2x + 16 + 2x = x^2 + 4x + 3^2 - 9\\ \\ &&& \mathbf{Exponents} \qquad \qquad \qquad \qquad \qquad \ \ 2x + 16 + 2x = x^2 + 4x + 9 - 9\\ \\ &&& \mathbf{Add \ variables} \qquad \qquad \qquad \qquad \quad 4x + 16 = x^2 + 4x + 9 - 9\\ \\ &&& \mathbf{Subtract} \ 4x'' \ \mathbf{from \ each \ side} \qquad 4x + 16 - 4x = x^2 + 4x + 9 - 9 - 4x\\ \\ &&& \mathbf{Add/Subtract} \qquad \qquad \qquad \qquad \quad 16 = x^2 + 9 - 9\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 16 = x^2\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4 = x\end{align*}

\begin{align*}&\mathbf{Check :} && \text{Replace} \ x \ \text{with} \ 4 \ \text{in the equation. Check for equality.}\\ &&& \qquad 2(4 + 8) + 2 \times 4 = 4(4 + 2^2) + 3^2 - 9\\ &&& \qquad 2 \times 12 + 2 \times 4 = 4 \times (4 + 4) + 9 - 9\\ &&& \qquad 24 + 8 = 32 + 9 - 9\\ &&& \qquad 32 = 32\end{align*}

Solve for the value of \begin{align*}x\end{align*}.

Show all steps.

\begin{align*}1. \quad x^2 + 4(3 + x) - 3 = 2(2x + 9)\\ \\ \\ \\ \\ 2. \quad 4(3 - x) + 9^2 = x(x - 4) + 7^2 - 5\\ \\ \\ \\ \\ 3. \quad 5(x + 8) + 8^2 - (5^2 + 3 \times 13) = x(x+5) + (9-7)^2 \\ \\ \\ \\ \end{align*}

Extra for Experts: \begin{align*}\textbf{\textit{X}}\end{align*} the Unknown – Solve for Unknowns using the Distributive Property

Solve for the value of \begin{align*}x\end{align*}.

Show all steps.

\begin{align*}1. \quad x^2 + 6(x - 1) = 2(3x - 2) + 2^2 - 2\\ \\ \\ \\ \\ 2. \quad x^2 + 4(x + 1) = 2(2x + 3^2) + 11\\ \\ \\ \\ \\ 3. \quad 3(x + 10) - 1 - 2^2 = 3x + x^2 + 2(3 + 5) \\ \\ \\ \\ \end{align*}

Solutions:

\begin{align*}1. \ \quad x^2 + 6(x - 1) = 2(3x - 2) + 2^2 - 2\!\\ {\;} \qquad x^2 + 6x - 6 = 6x - 4 + 2^2 - 2\!\\ {\;} \qquad x^2 + 6x - 6 = 6x - 4 + 4 - 2\!\\ {\;} \qquad x^2 + 6x - 6 - 6x = 6x - 4 + 4 - 2 - 6x\!\\ {\;} \qquad x^2 + -6 = - 4 + 4 - 2\!\\ {\;} \qquad x^2 + -6 = - 2\!\\ {\;} \qquad x^2 + -6 + 6 = - 2 + 6\!\\ {\;} \qquad x^2 = 4\!\\ {\;} \qquad x = 2\!\\ \\ 2. \quad \ x^2 + 4(x + 1) = 2(2x + 3^2) + 11\!\\ {\;} \qquad x^2 + 4x + 4 = 4x + 2 \times 9 + 11\!\\ {\;} \qquad x^2 + 4x + 4 - 4x = 4x + 2 \times 9 + 11 - 4x \!\\ {\;} \qquad x^2 + 4 = 2 \times 9 + 11\!\\ {\;} \qquad x^2 + 4 = 18 + 11\!\\ {\;} \qquad x^2 + 4 - 4 = 18 + 11 - 4\!\\ {\;} \qquad x^2 = 25\!\\ {\;} \qquad x = 5\!\\ \\ 3. \quad \ 3(x + 10) - 1 - 2^2 = 3x + x^2 + 2(3 + 5)\!\\ {\;} \qquad 3x + 30 - 1 - 2^2 = 3x + x^2 + 2 \times 8\!\\ {\;} \qquad 3x + 30 - 1 - 4 = 3x + x^2 + 2 \times 8\!\\ {\;} \qquad 3x + 30 - 1 - 4 - 3x = 3x + x^2 + 2 \times 8 - 3x \!\\ {\;} \qquad 30 - 1 - 4 = x^2 + 2 \times 8\!\\ {\;} \qquad 30 - 1 - 4 = x^2 + 16\!\\ {\;} \qquad 25 = x^2 + 16\!\\ {\;} \qquad 25 - 16 = x^2 + 16 - 16\!\\ {\;} \qquad 9 = x^2\!\\ {\;} \qquad 3 = x\end{align*}

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Date Created:
Feb 23, 2012
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May 14, 2015
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CK.MAT.ENG.SE.1.Algebra-Explorations-K-7.9.2