<meta http-equiv="refresh" content="1; url=/nojavascript/">

# 8.3: Solve for Unknowns

Difficulty Level: At Grade Created by: CK-12

Solve for Unknowns – Apply the Distributive Property

Teacher Notes

To solve for values of unknowns in equations, students perform all computations following the order of operations: operations in parentheses first, followed by exponentiation, multiplication and division left to right, then addition and subtraction left to right. When the variable is in parentheses, students must apply the distributive property of multiplication over addition: $a \times (b + c) = a \times b + a \times c$. Encourage students to check by replacing the variable in the equation with its value, perform the computations, and verify that the two expressions (to the left and to the right of the = symbol) name the same number.

Solutions:

$1. \quad \ (3 + 5)^2 \div 2 - 2 = 2b + 4^2\!\\{\;} \qquad 8^2 \div 2 - 2 = 2b + 4^2\!\\{\;} \qquad 64 \div 2 - 2 = 2b + 16\!\\{\;} \qquad 32 - 2 = 2b + 16\!\\{\;} \qquad 30 = 2b + 16\!\\{\;} \qquad 30 - 16 = 2b\!\\{\;} \qquad 14 = 2b\!\\{\;} \qquad 7 = b$

$2. \quad \ 4(c + 2) + 2 = 2 \times 5^2\!\\{\;} \qquad 4c + 8 + 2 = 2 \times 5^2\!\\{\;} \qquad 4c + 8 + 2 = 2 \times 25\!\\{\;} \qquad 4c + 8 + 2 = 50\!\\{\;} \qquad 4c + 10 = 50\!\\{\;} \qquad 4c = 50 - 10\!\\{\;} \qquad 4c = 40\!\\{\;} \qquad c = 10$

$3. \quad \ 2 + (16 - 4) \div 3 + d + 2d = 9^2 \div 9 \times 2\!\\{\;} \qquad 2 + 12 \div 3 + d + 2d = 9^2 \div 9 \times 2\!\\{\;} \qquad 2 + 12 \div 3 + d + 2d = 81 \div 9 \times 2\!\\{\;} \qquad 2 + 4 + d + 2d = 18\!\\{\;} \qquad 6 + 3d = 18\!\\{\;} \qquad 3d = 18 - 6\!\\{\;} \qquad 3d = 12\!\\{\;} \qquad d = 4$

$4. \quad \ 61 - (9 - 6)^2 = 5(2 + f) - 3^1\!\\{\;} \qquad 61 - 3^2 = 5 \times 2 + 5f - 3^1\!\\{\;} \qquad 61 - 9 = 5 \times 2 + 5f - 3\!\\{\;} \qquad 61 - 9 = 10 + 5f - 3\!\\{\;} \qquad 52 = 5f + 7\!\\{\;} \qquad 52 - 7 = 5f\!\\{\;} \qquad 45 = 5f\!\\{\;} \qquad 9 = f$

Solve for Unknowns – Apply the Distributive Property

$4 \times 1 + 6(a - 1) - 9 - 2^3 = 2 + 3^2$

What is the value of $a$?

$& \mathbf{Describe:} && \text{The} \ a - 1 \ \text{is in parentheses and is multiplied by 6. The 2 has an exponent of 3, and}\\&&&\text{the 3 has an exponent of 2. The other operations shown are multiplication,}\\&&&\text{addition, and subtraction.}\\& \mathbf{My \ Job:} && \text{Apply the distributive property and do the order of operations in order to solve for}\\&&&\text{the value of} \ a.\\& \mathbf{Solve:} && 4 \times 1 + 6(a - 1) - 9 - 2^3 = 2 + 3^2\\&&& \mathbf{Distributive} \qquad \qquad 4 \times 1 + 6a - 6 - 9 - 2^3 = 2 + 3^2\\&&& \mathbf{property}\\&&& \mathbf{Exponents} \qquad \qquad \quad 4 \times 1 + 6a - 6 - 9 - 8 = 2 + 9\\&&& \mathbf{Multiplication/} \qquad \ \ 4 + 6a - 6 - 9 - 8 = 2 + 9\\&&& \mathbf{Division}\\&&& \mathbf{(left \ to \ right)}\\&&& \mathbf{Addition} \qquad \qquad \qquad 6a - 19 = 11\\&&& \mathbf{(left \ to \ right)} \qquad \qquad 6a = 11 + 19\\&&& \qquad \qquad \qquad \qquad \qquad \ \ 6a = 30\\&&& \qquad \qquad \qquad \qquad \qquad \ \ a = 5\\& \mathbf{Check:} && \text{Replace} \ a \ \text{with 5 in the equation. Check that the two expressions (to the right and}\\&&&\text{to the left of the = symbol) name the same number.}\\&&& 4 \times 1 + 6(a - 1) - 9 - 2^3 = 2 + 3^2\\&&& 4 \times 1 + 6(5 - 1) - 9 - 2^3 = 2 + 3^2\\&&& 4 \times 1 + 6 \times 4 - 9 - 2^3 = 2 + 3^2\\&&& 4 \times 1 + 6 \times 4 - 9 - 8 = 2 + 9\\&&& 4 + 24 - 9 - 8 = 2 + 9\\&&& 11 = 11$

Solve for the unknown.

Follow the order of operations. Show each step.

$1. \quad (3 + 5)^2 \div 2 - 2 = 2b + 4^2\\\\\\\\\\{\;} \qquad b = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\2. \quad 4(c + 2) + 2 = 2 \times 5^2\\\\\\\\\\{\;} \qquad c = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\3. \quad 2 + (16 - 4) \div 3 + d + 2d = 9^2 \div 9 \times 2\\\\\\\\\\{\;} \qquad d = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\4. \quad 61 - (9 - 6)^2 = 5 (2 + f) - 3^1\\\\\\\\\\{\;} \qquad y = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

Extra for Experts: Solve for Unknowns – Apply the Distributive Property

Solve for the unknown.

Follow the order of operations. Show each step.

$1. \quad 5h + 3(h - 2) + 2 = 2^2 \times 3\\\\\\\\\\{\;} \qquad h = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\2. \quad 10^2 - 5 \times 8 = 1 + 6(j + 4) + (5^2 + 5) \div 6\\\\\\\\\\{\;} \qquad j = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\3. \quad k + 2(6 + k) + 3^2 = 2^3 \times 3 \times 2 - 3\\\\\\\\\\{\;} \qquad k = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\4. \quad 6^2 + 2^2 = 2(2m + 4m) - 4^2 \times 2\\\\\\\\\\{\;} \qquad m = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

Solutions:

$1. \quad \ 5h + 3(h - 2) + 2 = 2^2 \times 3\!\\{\;} \qquad 5h + 3h - 6 + 2 = 2^2 \times 3\!\\{\;} \qquad 5h + 3h - 6 + 2 = 4 \times 3\!\\{\;} \qquad 5h + 3h - 6 + 2 = 12\!\\{\;} \qquad 8h - 4 = 12\!\\{\;} \qquad 8h = 12 + 4\!\\{\;} \qquad 8h = 16\!\\{\;} \qquad h = 2\!\\\!\\2. \quad \ 10^2 - 5 \times 8 = 1 + 6 (j + 4) + (5^2 + 5) \div 6\!\\{\;} \qquad 10^2 - 5 \times 8 = 1 + 6 j + 24 + (25 + 5) \div 6\!\\{\;} \qquad 10^2 - 5 \times 8 = 1 + 6 j + 24 + 30 \div 6\!\\{\;} \qquad 100 - 5 \times 8 = 1 + 6 j + 24 + 30 \div 6\!\\{\;} \qquad 100 - 40 = 1 + 6 j + 24 + 5\!\\{\;} \qquad 60 = 6 j + 30\!\\{\;} \qquad 60 - 30 = 6 j\!\\{\;} \qquad 30 = 6 j\!\\{\;} \qquad 5 = j\!\\\\3. \quad \ k + 2(6 + k) + 3^2 = 2^3 \times 3 \times 2 - 3\!\\{\;} \qquad k + 12 + 2k + 3^2 = 2^3 \times 3 \times 2 - 3\!\\{\;} \qquad k + 12 + 2k + 9 = 8 \times 3 \times 2 - 3\!\\{\;} \qquad k + 12 + 2k + 9 = 48 - 3\!\\{\;} \qquad 3k + 21 = 45\!\\{\;} \qquad 3k = 45 - 21\!\\{\;} \qquad 3k = 24\!\\{\;} \qquad k = 8\!\\\\4. \quad \ 6^2 + 2^2 = 2(2m + 4m) - 4^2 \times 2\!\\{\;} \qquad 6^2 + 2^2 = 2 \times 6m - 4^2 \times 2\!\\{\;} \qquad 36 + 4 = 2 \times 6m - 16 \times 2\!\\{\;} \qquad 36 + 4 = 12m - 32\!\\{\;} \qquad 40 = 12m - 32\!\\{\;} \qquad 40 + 32 = 12m\!\\{\;} \qquad 72 = 12m\!\\{\;} \qquad 6 = m$

## Date Created:

Feb 23, 2012

Nov 05, 2014
You can only attach files to None which belong to you
If you would like to associate files with this None, please make a copy first.