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Step-4

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	\$ 14/5 t + 14 = 4/5 t - 6 \$
2	Move Variable terms to the left side and constant terms to the right side	\$ ((14t)/5) + ((4t)/5) = -6 - 14 \$
3	Taking LCM as '\$5\$'	\$ (14t + 4t)/5 = -20/1\$
4	By using cross multiplication	\$ 14t + 4t = -100\$
5	After Addition	\$ 18t = -100 \$
6	Dividing on both sides by '\$18\$' of the equality	\$ \cancel(18)^1/ \cancel(18)^1 t = -\cancel100^(50)/\cancel18^9 \$
7	After cancelation	\$ t = - 50/9 \$
8	Answer	B = \$ t = - 50/9 \$

Steps

Problem

Solution

Solve

\$ 14/5 t + 14 = 4/5 t - 6 \$

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

\$ ((14t)/5) + ((4t)/5) = -6 - 14 \$

Taking LCM as '\$5\$'

\$ (14t + 4t)/5 = -20/1\$

By using cross multiplication

\$ 14t + 4t = -100\$

After Addition

\$ 18t = -100 \$

Dividing on both sides by '\$18\$' of the equality

\$ \cancel(18)^1/ \cancel(18)^1 t = -\cancel100^(50)/\cancel18^9 \$

After cancelation

\$ t = - 50/9 \$

Answer

B = \$ t = - 50/9 \$