We want to isolate the k, which is being divided by 10. The first thing we should do is multiply both sides by 10.

 

[latex]\begin{array}{l}\text{ Multiply the left and right sides of the equation by 10 }\end{array}[/latex]

[latex](10)(-\frac{7}{2})=\frac{k}{10}(10)[/latex]

[latex]\begin{array}{l}\text{ Simplify each side }\end{array}[/latex]

[latex]\frac{-70}{2} = \frac{k\cdot10}{10}[/latex]

[latex]\begin{array}{l}\text{ Simplify again }\end{array}[/latex]

[latex]-35 =k[/latex]

Answer

We write the k on the left side as a matter of convention.

[latex]k=-35[/latex]

Notice the least common multiple for the denominators of 2 and 4 is 4. We will now use this LCM to clear fractions.

Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.

[latex]\begin{array}{r}\frac{1}{2}x-3=2-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\end{array}[/latex]

Use the distributive property to expand the expressions on both sides. Multiply.

[latex]\begin{array}{r}4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)-4\left(\frac{3}{4}x\right)\\\\ \frac{4}{2}x-12=8-\frac{12}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}[/latex]

Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.

[latex]\begin{array}{r}2x-12=8-3x\, \\\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Add 12 to both sides to move the constant terms to the other side.

[latex]\begin{array}{r}5x-12=8\,\,\\ \underline{\,\,\,\,\,\,+12\,+12} \\5x=20\end{array}[/latex]

Divide to isolate the variable.

[latex]\begin{array}{r}\underline{5x}=\underline{20}\\ 5\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\\ x=4\end{array}[/latex]

Answer

[latex]x=4[/latex]

Find the least common denominator of [latex]4[/latex] and [latex]8[/latex]. Remember, to find the LCD, identify the greatest number of times each factor appears in each factorization. Here, [latex]2[/latex] appears [latex]3[/latex] times, so [latex]2\cdot2\cdot2[/latex], or [latex]8[/latex], will be the LCD.

Multiply both sides of the equation by the common denominator, [latex]8[/latex], to keep the equation balanced and to eliminate the denominators.

[latex]\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}[/latex]

Simplify and solve for x.

[latex]\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}[/latex]

Check the solution by substituting [latex]9[/latex] for x in the original equation.

[latex]\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}[/latex]

Therefore, [latex]x=9[/latex].

Find the least common denominator of [latex]5[/latex] and [latex]3[/latex].  Here, [latex]5[/latex] and [latex]3[/latex] don’t have any common factors so  [latex]5\cdot3=15[/latex], will be the LCD.

Multiply both sides of the equation by the common denominator, [latex]15[/latex], to keep the equation balanced and to eliminate the denominators.

[latex]\begin{array}{r}15\cdot \frac{x+3}{5}=\frac{x+8}{3}\cdot 15\,\,\,\,\,\,\,\\\\\frac{15(x+3)}{5}=\frac{(x+8)15}{3}\,\,\,\,\,\,\\\\\frac{15}{5}\cdot (x+3)=(x+8)\cdot\frac{15}{3}\\\\3\cdot (x+3)=(x+8)\cdot 5\\\\3x+9=5x+40,\,\,\end{array}[/latex]

Simplify and start to solve for x by collecting x terms to one side.

[latex]\begin{array}{r}3x+9=5x+40\,\\\underline{-5x\,\,\,\,\,\,\,\,\,\,\,\,-5x\,\,\,\,\,\,\,\,\,\,\,\,}\\ -2x+9=\,\,\,\,\,\,\,\,\,\,\,40\,\,\,\end{array}[/latex]

Subtract 9 on both sides to move constants to the other side of the equal sign.

[latex]\begin{array}{r}-2x+9=40\,\,\\ \underline{\,\,\,\,\,\,-9\,\,\,\,-9}\, \\-2x=31\end{array}[/latex]

Divide by -2 on both sides to isolate the x term.

[latex]\begin{array}{r}\frac{-2x}{-2}=\frac{31}{-2}\end{array}[/latex]

[latex]x=-\frac{31}{2}=-15.5[/latex]

Check the solution by substituting [latex]-15.5[/latex] for x in the original equation.

[latex]\begin{array}{r}\frac{x+3}{5}=\frac{x+8}{3}\\\\\frac{-15.5+3}{5}=\frac{-15.5+8}{3}\\\\\frac{-12.5}{5}=\frac{-7.5}{3}\\\\2.5=2.5\end{array}[/latex]

Therefore, [latex]x=-15.5[/latex].