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Step-1

Title: Solve the equation

Grade: 7-b Lesson: S2-L6

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Steps	Problem	Solution
1	Solve	\$ x/2 - 4 = (5x)/2 - 14 \$
2	Move Variable terms to the left side and constant terms to the right side	\$ (x/2) + ((5x)/2) = -14 + 4 \$
3	Taking LCM as '\$2\$'	\$ (x + 5x)/2 = -10/1\$
4	By using cross multiplication	\$ x + 5x = -20\$
5	After Addition	\$ 6x = -20 \$
6	Dividing on both sides by '\$6\$' of the equality	\$ \cancel(6)^1/ \cancel(6)^1 x = -\cancel20^(10)/\cancel6^3 \$
7	After cancelation	\$ x = - 10/3 \$
8	Answer	D = \$ x = - 10/3 \$

Steps

Problem

Solution

Solve

\$ x/2 - 4 = (5x)/2 - 14 \$

Move Variable terms to the left side and constant terms to the right side

\$ (x/2) + ((5x)/2) = -14 + 4 \$

Taking LCM as '\$2\$'

\$ (x + 5x)/2 = -10/1\$

By using cross multiplication

\$ x + 5x = -20\$

After Addition

\$ 6x = -20 \$

Dividing on both sides by '\$6\$' of the equality

\$ \cancel(6)^1/ \cancel(6)^1 x = -\cancel20^(10)/\cancel6^3 \$

After cancelation

\$ x = - 10/3 \$

Answer

D = \$ x = - 10/3 \$