Solving Equations by using Addition/ Subtraction: Solving an equation with fractions

Solve for X:

$\frac{1}{8}x-\frac{2}{3}x+\frac{1}{5}x=1$

The common denominator of 8, 3, and 5 is 120.

Now we multiply each numerator by the corresponding number to reach 120 and thus cancel the fractions and obtain the following equation:

$(1\times x\times15)-(2\times x\times40)+(1\times x\times24)=1\times120$

We multiply the exercises in parentheses accordingly:

$15x-80x+24x=120$

We will solve the left side (from left to right) and will obtain:

$(15x-80x)+24x=120$

$-65x+24x=120$

$-41x=120$

We reduce both sides by $-41$

$\frac{-41x}{-41}=\frac{120}{-41}$

We find that x is equal$x=-\frac{120}{41}$

$-\frac{120}{41}$

$a+2\frac{1}{2}=4$

$a=1\frac{1}{2}$

Solve for X:

$\frac{2}{7}x-\frac{2}{3}x=4$

$10\frac{1}{2}$

Solve for X:

$\frac{2}{5}x+\frac{3}{4}x=1$

$\frac{20}{23}$

Solve for X:

$\frac{1}{8}x+\frac{2}{3}x=3$

$3\frac{15}{19}$

Solve for X:

$\frac{1}{3}x-\frac{1}{2}x=3$

$-18$

$b-3\frac{4}{5}=-8\frac{3}{5}$

$b=-4\frac{4}{5}$

Solve for X:

$\frac{1}{4}x-\frac{2}{3}x+\frac{1}{8}x=5$

$-17\frac{1}{7}$

Solve for X:

$-\frac{1}{4}x+\frac{2}{3}x-\frac{1}{6}x=2$

$8$

Solve for X:

$\frac{2}{3}x+\frac{1}{4}x-\frac{1}{5}x+\frac{1}{2}x=2$

$\frac{120}{73}$

Solve for X:

$\frac{1}{4}x-\frac{1}{5}x+\frac{2}{3}x-\frac{2}{5}x=1$

$\frac{15}{7}$

Solve for X:

$\frac{2}{4}x+\frac{1}{2}x+\frac{3}{8}x-\frac{1}{5}x=1$

$\frac{40}{47}$