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QuestionWhich equation has a solution of t=5t=5? Choose one: a. 2(2t−4)−2t=−22(2 t-4)-2 t=-2 b. 2−3(t+4)=2t+152-3(t+4)=2 t+15 c. 8t−10+2t=5t−358 t-10+2 t=5 t-35 d. 3t+10=6t−53 t+10=6 t-5

Studdy Solution

STEP 1

Assumptions1. We are given four equations and we need to find out which one has a solution of t=5t=5.

STEP 2

We will start by substituting t=5t=5 into each equation and see which one holds true.

STEP 3

Substitute t=5t=5 into equation a.
2(2×5−)−2×5=−22(2 \times5 -) -2 \times5 = -2

STEP 4

implify the equation.
2(10−4)−10=−22(10 -4) -10 = -2

STEP 5

Calculate the result.
2()−10=−22() -10 = -2

STEP 6

Calculate the result.
12−10=−212 -10 = -2

STEP 7

Calculate the result.
2≠−22 \neq -2Equation a does not hold true for t=5t=5.

STEP 8

Substitute t=5t=5 into equation b.
2−3(5+4)=2×5+152 -3(5 +4) =2 \times5 +15

STEP 9

implify the equation.
2−3(9)=+152 -3(9) = +15

STEP 10

Calculate the result.
2−27=252 -27 =25

STEP 11

Calculate the result.
−25≠25-25 \neq25Equation b does not hold true for t=5t=5.

STEP 12

Substitute t=5t=5 into equation c.
8×5−10+2×5=5×5−358 \times5 -10 +2 \times5 =5 \times5 -35

STEP 13

implify the equation.
40−10+10=25−3540 -10 +10 =25 -35

STEP 14

Calculate the result.
40=−1040 = -10Equation c does not hold true for t=t=.

STEP 15

Substitute t=5t=5 into equation d.
3×5+10=×5−53 \times5 +10 = \times5 -5

STEP 16

implify the equation.
15+10=30−515 +10 =30 -5

STEP 17

Calculate the result.
25=2525 =25Equation d holds true for t=5t=5.
The equation that has a solution of t=5t=5 is equation d 3t+10=6t−53 t+10=6 t-5.

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