Question

How do I do these problems?

Answer 1

12) x= 21

13) x = 1

14) x = 9

15) x = 1

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:

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

Check your work:

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

13) Solve for x:

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

Check your work:

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

14) Solve for x:

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

`\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}`

Check your work:

`\sf 14)\ (x-1)/(6) - (x+1)/(8) = (1)/(12)\ \textsf{[substitute 9 for the value of x]}\\\\\implies (9-1)/(6) - (9+1)/(8) = (1)/(12)\ \textsf{[simplify]}\\\\\implies (8)/(6) - (10)/(8) = (1)/(12)\ \textsf{[rewrite the fractions with a common denominator of 24]}\\\\\implies (8*4)/(6*4) - (10*3)/(8*3) = (1*2)/(12*2)\ \textsf{[simplify]}\\\\`

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

15) Solve for x:

`\sf (1)/(4)+(x+1)/(8)=(1)/(2)\ \textsf{[multiply both sides by 8 - LCM of 2, 4, and 8]}\\\\\implies 8\left((1)/(4)+(x+1)/(8)\right)=8\left((1)/(2)\right)\textsf{[multiply]}\\\\\implies \left((8)/(4)+(8x+8)/(8)\right)=(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}`

Check your work:

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