Respuesta :

lublana

Given problem is


[tex]\frac{2}{3}w-\frac{1}{4}=\frac{2}{3}\left(w-\frac{1}{4}\right)[/tex]


Here unknown variable is w so we will solve this equation for w.

Common denominator for given denominators 3 and 4 is 12 so we can multiply both sides by 12 so that we can cancel out all the fractions


[tex]\frac{2}{3}w\cdot12-\frac{1}{4}\cdot12=\frac{2}{3}\cdot12\left(w-\frac{1}{4}\right)[/tex]


[tex]\frac{24}{3}w-\frac{12}{4}=\frac{24}{3}\left(w-\frac{1}{4}\right)[/tex]



[tex]8w-3=8\left(w-\frac{1}{4}\right)[/tex]


[tex]8w-3=8w-\frac{8}{4}[/tex]


[tex]8w-3=8w-2[/tex]

[tex]8w-8w=3-2[/tex]


[tex]0=1[/tex]

We know that 0=1 is not possible

Hence there is no solution for w.

alinakincsem

Answer:

No solution

Step-by-step solution:

We are given an expression:

[tex]\frac{2}{3} w-\frac{1}{4} =\frac{2}{3} (w-\frac{1}{4} )[/tex]

We have only one variable [tex]w[/tex] so we will make [tex]w[/tex] the subject and solve for it.

[tex]\frac{2}{3} w-\frac{1}{4} =\frac{2}{3} (w-\frac{1}{4} )[/tex]

Solving the brackets first to get:

[tex]\frac{2}{3} w -\frac{1}{4} =\frac{2}{3} -\frac{1}{6}[/tex]

Taking LCM on both sides to get:

[tex]\frac{8w-3}{12} =\frac{4w-3}{6}[/tex]

By cross multiplication:

[tex]6(8w-3)=12(4w-3)[/tex]

[tex]48w-18 = 48w -36[/tex]

[tex]48w - 49w -18 + 36 = 0[/tex]

[tex]w[/tex] gets cancelled so we are left with [tex]18\neq 0[/tex]

Therefore, [tex]w[/tex] has no solution.

Otras preguntas