Question

5(x-3) =(x+2)
Can you please help me ​

Answer 1

Given,

5(x-3)=3(x-2)

5x-15=3x-6

5x-3x=15-6

2x=9

x=9/2

x=4.5

Answer 2

★ How to do :-

Here, we are given with some constants and two variables. We are asked to find the value of that variable x. Here, we are going to shift the numbers and the variables into one side and the constants on one side and then, we can find the value of that variable x. We should remember that, while shifting the numbers from one side to another we should change the sign of the particular number or variable. We also verify the statement by applying the value of x. So, let's solve!!

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➤ Solution :-

${\tt \leadsto 5 \: (x - 3) = (x + 2)}$

Multiply the number 5 with both numbers in the bracket.

${\tt \leadsto (5x - 15) = (x + 2)}$

Shift the variable on RHS to LHS and the constant on LHS to RHS, changing it's sign.

${\tt \leadsto 5x - x = 15 + 2}$

Subtract the variables on LHS and the constants on RHS.

${\tt \leadsto 4x = 17}$

Shift the number 4 from LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{17}{4}}$

Simplify the fraction to get the value of x.

${\tt \leadsto \pink{\underline{\boxed{\tt x = 4.25}}}}$

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Verification :-

${\tt \leadsto 5 \: (x - 3) = (x + 2)}$

Substitute the value of x.

${\tt \leadsto 5 \: (4.25 - 3) = (4.25 + 2)}$

Add amd subtracting to the thr numbers in bracket.

${\tt \leadsto 5 \: (1.25) = 6.25}$

Multiply the numbers on LHS.

${\tt \leadsto 6.25 = 6.25}$

So,

${\tt \leadsto LHS = RHS}$

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Hence verified !!