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# 9.1: X for Unknown

Difficulty Level: At Grade Created by: CK-12

Look at the equation below. Can you figure out the value of x\begin{align*}x\end{align*}? In this concept, you will learn how to solve equations using the order of operations, distributive property, and square roots.

2(x+8)+2x=x(x+22)+329\begin{align*}2(x + 8) + 2x = x(x + 2^2) + 3^2 - 9\end{align*}

### Guidance

The order of operations tells us the correct order of evaluating math expressions. We always do parentheses first. Then we do exponents. Next we do multiplication and division (from left to right) and finally addition and subtraction (from left to right). The distributive property (a×(b+c)=a×b+a×c)\begin{align*}(a \times (b + c) = a \times b + a \times c)\end{align*} allows us to remove parentheses when there is an unknown inside of them.

In order to evaluate expressions using the distributive property and the order of operations, we can use the problem solving steps to help.

• First, describe what you see in the problem. What operations are there?
• Second, identify what your job is. In these problems, your job will be to solve for the unknown.
• Third, make a plan. In these problems, your plan should be to use the distributive property and the order of operations to get the x2\begin{align*}x^2\end{align*} by itself. At the end, you will use a square root.
• Fourth, solve.
• Fifth, check. Substitute your answer into the equation and make sure it works.

#### Example A

Solve for the positive value of x\begin{align*}x\end{align*}. Show all steps.

x2+4(3+x)3=2(2x+9)\begin{align*} x^2 + 4(3 + x) - 3 = 2(2x + 9)\end{align*}

Solution:

We can use problem solving steps to help us to solve this equation.

\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 x^2 + 4(3 + x) - 3 = 2(2x + 9)\\ &&& \mathbf{Parentheses} \qquad \qquad \quad x^2 + 12+4x - 3 = 4x+18\\ \\ &&& \mathbf{Combine \ like \ terms} \qquad \qquad \quad x^2 + 9+4x = 4x+18\\ \\ &&& \mathbf{Add } \ 4x \ \mathbf{to \ each \ side} \qquad x^2 + 9+4x-4x = 4x+18-4x\\ \\ &&& \mathbf{Add/Subtract} \qquad \qquad \qquad \qquad \quad x^2+9=18\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x^2 = 9\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x = 3\end{align*}

\begin{align*}&\mathbf{Check :} && \text{Replace} \ x \ \text{with} \ 3 \ \text{in the equation. Check for equality.}\\ &&& \qquad 3^2 + 4(3 + 3) - 3 = 2(2 \times 3 + 9)\\ &&& \qquad 3^2 + 4 \times 6 - 3 = 2(6 + 9)\\ &&& \qquad 3^2 + 4 \times 6 - 3 = 2 \times 15\\ &&& \qquad 9+ 4 \times 6 - 3 = 2 \times 15\\ &&& \qquad 9+ 24 - 3 = 30\\ &&& \qquad 30=30\end{align*}

#### Example B

Solve for the positive value of \begin{align*}x\end{align*}. Show all steps.

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

Solution:

We can use problem solving steps to help us to solve this equation.

\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 4(3 - x) + 9^2 = x(x - 4) + 7^2 - 5\\ &&& \mathbf{Parentheses} \qquad \qquad \quad 12-4x + 9^2 = x^2-4x + 7^2 - 5\\ \\ &&& \mathbf{Exponents} \qquad \qquad \qquad \qquad \ \ 12-4x + 81 = x^2-4x + 49 - 5\\ \\ &&& \mathbf{Combine \ like \ terms} \qquad \qquad \quad 93-4x = x^2-4x + 44\\ \\ &&& \mathbf{Add} \ 4x \ \mathbf{to \ each \ side} \qquad 93-4x+4x = x^2-4x+4x + 44\\ \\ &&& \mathbf{Add/Subtract} \qquad \qquad \qquad \qquad \quad 93 = x^2 + 44\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 49 = x^2\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 7 = x\end{align*}

\begin{align*}&\mathbf{Check :} && \text{Replace} \ x \ \text{with} \ 7 \ \text{in the equation. Check for equality.}\\ &&& \qquad 4(3 - 7) + 9^2 = 7(7 - 4) + 7^2 - 5\\ &&& \qquad 4 \times -4 + 9^2 = 7 \times 3 + 7^2 - 5\\ &&& \qquad 4 \times -4 + 81 = 7 \times 3 + 49 - 5\\ &&& \qquad -16 + 81 = 21 + 49 - 5\\ &&& \qquad 65=65\end{align*}

#### Example C

Solve for the positive value of \begin{align*}x\end{align*}. Show all steps.

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

Solution:

We can use problem solving steps to help us to solve this equation.

\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 5(x + 8) + 8^2 - (5^2 + 3 \times 13) = x(x+5) + (9-7)^2\\ &&& \mathbf{Parentheses} \qquad \qquad \quad 5x+40 + 8^2 - (25+39) = x^2+5x + 2^2\\ \\ &&& \mathbf{More \ Parentheses} \qquad \qquad \quad 5x+40 + 8^2 - 64 = x^2+5x + 2^2\\ \\ &&& \mathbf{Exponents} \qquad \qquad \qquad \qquad \qquad \ \ 5x+40 + 64 - 64 = x^2+5x + 4\\ \\ &&& \mathbf{Combine \ like \ terms} \qquad \qquad \qquad \qquad \quad 5x+40 = x^2+5x + 4\\ \\ &&& \mathbf{Subtract} \ 5x \ \mathbf{from \ each \ side} \qquad 5x+40 -5x= x^2+5x + 4-5x\\ \\ &&& \mathbf{Add/Subtract} \qquad \qquad \qquad \qquad \quad 40 = x^2 + 4\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 36 = x^2\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 6 = x\end{align*}

\begin{align*}&\mathbf{Check :} && \text{Replace} \ x \ \text{with} \ 6 \ \text{in the equation. Check for equality.}\\ &&& \qquad 5(6 + 8) + 8^2 - (5^2 + 3 \times 13) = 6(6+5) + (9-7)^2\\ &&& \qquad 5 \times 14 + 8^2 - (25+39) = 6 \times 11 + 2^2\\ &&& \qquad 5 \times 14 + 8^2 - 64 = 6 \times 11 + 2^2\\ &&& \qquad 5 \times 14 +64-64=6 \times 11 +4\\ &&& \qquad 70 +64-64=66 +4\\ &&& \qquad 70 = 70\end{align*}

#### Concept Problem Revisited

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

We can use problem solving steps to help us to solve this equation.

\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*}

### Vocabulary

The order of operations tells us the correct order of evaluating math expressions. We always do parentheses first. Then we do exponents. Next we do multiplication and division (from left to right) and finally addition and subtraction (from left to right). The distributive property \begin{align*}(a \times (b + c) = a \times b + a \times c)\end{align*} allows us to remove parentheses when there is an unknown inside of them. The square root of a number is a value that, when multiplied by itself, gives the original number.

### Guided Practice

Solve for the positive value of \begin{align*}x\end{align*}. Show all steps.

1. \begin{align*} x^2 + 6(x - 1) = 2(3x - 2) + 2^2 - 2\end{align*}

2. \begin{align*} x^2 + 4(x + 1) = 2(2x + 3^2) + 11\end{align*}

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

1. \begin{align*}x=2\end{align*}. Here are the steps to solve:

\begin{align*}\ \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\end{align*}

2. \begin{align*}x=5\end{align*}. Here are the steps to solve:

\begin{align*}\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\end{align*}

3. \begin{align*}x=3\end{align*}. Here are the steps to solve:

\begin{align*} \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*}

### Practice

Solve for the positive value of \begin{align*}x\end{align*}. Show all steps.

1. \begin{align*}x^2-4(x-3)=4(2x-1)+3^2-1-4x\end{align*}
2. \begin{align*}2x^2+7(x-5)=3x+4(x-9)+2^2+x^2+1\end{align*}
3. \begin{align*}x^2+5(2x+1)=2(5x+2^2)+22\end{align*}
4. \begin{align*}6(x-5)+13+3^2=6x+x^2-4(1+2)\end{align*}
5. \begin{align*}x^2-4(2x+6)=1-8x\end{align*}
6. \begin{align*}2(4x-3)+4^2-12=10x+x^2-2(1+x)\end{align*}
7. \begin{align*}3x^2+4(x-3)=2(2x+x^2)+24\end{align*}
8. \begin{align*}2(x-5)+16-2^3=x^2-7x+9(x-3)\end{align*}

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Jan 18, 2013