How do I do these problems?

Answer:

12) x= 21

13) x = 1

14) x = 9

15) x = 1

Step-by-step explanation:

Here, we are being asked to solve the given equations and check our work.

To check the work, we need to substitute the found value of x into the equation to check if it makes a true and correct statement.

12) Solve for x:

[tex]\sf 12)\ \dfrac{2}{3}x + 2 = 16\ \textsf{[subtract 2 from both sides]}\\\\\implies \dfrac{2}{3}x + 2 - 2 = 16 - 2\\\\\implies \dfrac{2}{3}x = 14\ \textsf{[multiply both sides by 3]}\\\\\implies 3\left(\dfrac{2}{3}\right)x = 14(3)\\\\\implies 2x = 42\ \textsf{[divide both sides by 2]}\\\\\implies \dfrac{2x}{2}=\dfrac{42}{2}\\\\\implies \boxed{\sf x = 21}[/tex]

Check your work:

[tex]\sf \dfrac{2}{3}x + 2 = 16\ \textsf{[substitute 21 for the value of x]}\\\\\implies \dfrac{2}{3}(21) + 2 = 16\ \textsf{[multiply]}\\\\\implies \dfrac{42}{3} + 2 = 16\ \textsf{[divide]}\\\\\implies 14 + 2 = 16\ \textsf{[add]}\\\\\implies 16 = 16\ \checkmark \textsf{[true statement]}[/tex]

13) Solve for x:

[tex]\sf 13)\ \dfrac{x}{2} + \dfrac{x}{3} = \dfrac{5}{6}\ \textsf{[multiply both sides by 6 - LCM of 2, 3, and 6]}\\\\\implies 6\left(\dfrac{x}{2} + \dfrac{x}{3}\right) = 6\left(\dfrac{5}{6}\right)\ \textsf{[multiply]}\\\\\implies \left(\dfrac{6x}{2} + \dfrac{6x}{3}\right) = \dfrac{30}{6}\ \textsf{[divide]}\\\\\implies 3x+2x=5\ \textsf{[combine like terms]}\\\\\implies 5x = 5\ \textsf{[divide both sides by 5]}\\\\\implies \dfrac{5x}{5}=\dfrac{5}{5}\\\\\implies \boxed{\sf x = 1}[/tex]

Check your work:

[tex]\sf \dfrac{x}{2} + \dfrac{x}{3} = \dfrac{5}{6}\ \textsf{[substitute 1 for the value of x]}\\\\\implies \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}\ \textsf{[rewrite the fractions with a common denominator of 6]}\\\\\implies \dfrac{1\times3}{2\times3} + \dfrac{1\times2}{3\times2} = \dfrac{5}{6}\ \textsf{[multiply]}\\\\\implies \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}\ \textsf{[add]}\\\\\implies \dfrac{5}{6} = \dfrac{5}{6}\ \checkmark \ \textsf{[true statement]}[/tex]

14) Solve for x:

[tex]\sf 14)\ \dfrac{x-1}{6} - \dfrac{x+1}{8} = \dfrac{1}{12}\ \textsf{[multiply both sides by 24 - LCM of 6, 8, and 24]}\\\\\implies 24\left(\dfrac{x-1}{6} - \dfrac{x+1}{8}\right) = 24\left(\dfrac{1}{12}\right)\ \textsf{[multiply]}\\\\\implies \left(\dfrac{24x-24}{6}\right) - \left(\dfrac{24x+24}{8}\right) = \dfrac{24}{12}\ \textsf{[divide]}\\\\[/tex]

[tex]\sf\\\implies (4x-4)-(3x+3)=2\ \textsf{[distribute the negative sign]}\\\\\implies (4x-4)+(-3x-3)=2\ \textsf{[combine like terms]}\\\\\implies x - 7 = 2\ \textsf{[add 7 to both sides]}\\\\\implies x - 7 + 7 = 2 + 7\\\\\implies \boxed{\sf x = 9}[/tex]

Check your work:

[tex]\sf 14)\ \dfrac{x-1}{6} - \dfrac{x+1}{8} = \dfrac{1}{12}\ \textsf{[substitute 9 for the value of x]}\\\\\implies \dfrac{9-1}{6} - \dfrac{9+1}{8} = \dfrac{1}{12}\ \textsf{[simplify]}\\\\\implies \dfrac{8}{6} - \dfrac{10}{8} = \dfrac{1}{12}\ \textsf{[rewrite the fractions with a common denominator of 24]}\\\\\implies \dfrac{8\times4}{6\times4} - \dfrac{10\times3}{8\times3} = \dfrac{1\times2}{12\times2}\ \textsf{[simplify]}\\\\[/tex]

[tex]\sf\\\implies \dfrac{32}{24}-\dfrac{30}{24}=\dfrac{2}{24}\ \textsf{[subtract]}\\\\\implies \dfrac{2}{24}=\dfrac{2}{24}\ \textsf{[reduce]}\\\\\implies \dfrac{1}{12}=\dfrac{1}{12}\ \checkmark\ \textsf{[true statement]}[/tex]

15) Solve for x:

[tex]\sf \dfrac{1}{4}+\dfrac{x+1}{8}=\dfrac{1}{2}\ \textsf{[multiply both sides by 8 - LCM of 2, 4, and 8]}\\\\\implies 8\left(\dfrac{1}{4}+\dfrac{x+1}{8}\right)=8\left(\dfrac{1}{2}\right)\textsf{[multiply]}\\\\\implies \left(\dfrac{8}{4}+\dfrac{8x+8}{8}\right)=\dfrac{8}{2}\ \textsf{[divide]}\\\\\implies 2 + x + 1 = 4\ \textsf{[add]}\\\\\implies x + 3 = 4\ \textsf{[subtract 3 from both sides]}\\\\\implies x + 3 - 3 = 4 - 3\\\\\implies \boxed{\sf x = 1}[/tex]

Check your work:

[tex]\sf \dfrac{1}{4}+\dfrac{x+1}{8}=\dfrac{1}{2}\ \textsf{[substitute 1 for the value of x]}\\\\\implies \left(\dfrac{1}{4}+\dfrac{1+1}{8}\right)=\dfrac{1}{2}\ \textsf{[simplify]}\\\\\implies \left(\dfrac{1}{4}+\dfrac{2}{8}\right)=\dfrac{1}{2}\ \textsf{[reduce]}\\\\\implies \left(\dfrac{1}{4}+\dfrac{1}{4}\right)=\dfrac{1}{2}\ \textsf{[add]}\\\\\implies \dfrac{2}{4}=\dfrac{1}{2}\ \textsf{[reduce]}\\\\\implies \dfrac{1}{2}=\dfrac{1}{2}\ \checkmark\ \textsf{[true statement]}[/tex]

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