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31 votes
31 votes
Consider the equation 5 – 3(2x – 7) = 12 – 5(x – 2). Is x = 2 a solution to this

equation? Is x = 4 a solution to this equation?

User Miche
by
2.5k points

2 Answers

14 votes
14 votes
Answer: No (x = 2), Yes (x = 4)

Explanation:

=> 5 - 3(2x - 7) = 12 - 5(x - 2)

Substitute x = 2 in the Equation :-

5 - 3(2x - 7) = 12 - 5(x - 2)
5 - 3(2(2) - 7) = 12 - 5((2) - 2)
5 - 3(4 - 7) = 12 - 5(2 - 2)
5 - 3(-3) = 12 - 5
5 + 9 = 12 - 5
14 = 7
LHS is not Equal to RHS

Therefore, x = 2 is not a Solution.

Substitute x = 4 in the Equation :-

5 - 3(2x - 7) = 12 - 5(x - 2)
5 - 3(2(4) - 7) = 12 - 5((4) - 2)
5 - 3(8 - 7) = 12 - 5(4 - 2)
5 - 3(1) = 12 - 5(2)
5 - 3 = 12 - 10
2 = 2
LHS is Equal to RHS

Therefore, x = 4 is a Solution.
User Reden
by
3.0k points
11 votes
11 votes

Answer:


\sf x=4

Explanation:


\sf 5 - 3(2x - 7)=12 - 5(x - 2)

To get rid of the parentheses, we'll apply the distributive property. Staring with the left side, multiply -3 by (2x-7).


\sf 5 - 6x+21=12 - 5(x - 2)

5+ 21= 26


\sf 26 - 6x=12 - 5(x - 2)

Now, let's do the same thing to the right side. Multiply -5 by (x-2) »12−5x+10, than add 12+ 10 = 22 .


\sf 26 - 6x=22 - 5x

Combine like terms.


\sf 26 - x=22

Subtract 26 from both sides.


\sf - x= - 4

Multiply both sides by -1.


\sf x=4

User Zvone
by
2.8k points
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