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# find a number such that 10 less than two-thirds the number is one-forth the number

I need to know how to solve it and the answer. this includes the equation(s) with the variable

I could be wrong, but if the number is x, then my best guesstimate would be...

x(2/3)-10=x(1/4)

Solve for x either as a fraction or a decimal.

Divide by x on both sides...

(2/3)-(10/x)=(1/4)

(-10/x)=(1/4)-(2/3)

(-10/x)=-0.4167

Divide both sides by -10...

(1/x)=0.04167

Take the inverse of both sides...

x=(1/0.04167)

x=24

Check your answer by substituting 24 for x in the original equation...

24(2/3)-10=24(1/4)

(8)(2)-10=(6)(1)

16-10=6, or...
6=6, TRUE

Therefore, x=24.

Make sense?

let x= the number
(2/3)x-10=(1/4)x
(2/3)x-(1/4)x=10
(8/12)x-(3/12)x=10
(5/12)x=10
(12/5)(5/12)x=(12/5)(10)
x=120/5
x=24
check: (2/3)(24)-10=(1/4)(24)
16-10=6
6=6

Hi Izzy;
[(2/3)x]-10=(1/4)(x)
Let's eliminate the fractions.
The denominators are 3 and 4.
(3)(4)=12
Let's multiply both sides by 12...
12{[(2/3)x]-10}=(1/4)(x)(12)
8x-120=3x
Let's subtract 8x from both sides...
-8x+8x-120=3x-8x
-120=-5x
Let's divide both sides by -5...
-120/-5=(-5x)/-5
24=x

Let's check our results...

[(2/3)x]-10=(1/4)(x)
[(2/3)(24)]-10=(1/4)(24)
16-10=6
6=6