Years
>
Year 8>
Equations>
Solve Equations with Variables on Both Sides using the Distributive Property>
Solve Equations with Variables on Both Sides using the Distributive Property 3Solve Equations with Variables on Both Sides using the Distributive Property 3
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
Question 1 of 4
1. Question
Solve`-4(2y-3)=1/2(12-4y)`- `y=` (1)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Distributive Property
`a``(b+c)=``a``b+``a``c`To solve for `y`, it needs to be alone on one side.Start by expanding the left side of the equation by using the Distributive Property.`-4``(2``y` `-3)` `=` `1/2``(12-4``y``)` `-4``(2``y``)` `-4``(-3)` `=` `1/2``(12)+``1/2``(-4``y``)` `-8``y` `+12` `=` `6-2``y` Next, move `-2y` to the other side by adding `2y` to both sides of the equation.`-8``y` `+12` `=` `6-2``y` `-8``y` `+12` `+2y` `=` `6-2``y` `+2y` `-6``y` `+12` `=` `6` `-2y+2y` cancels out Now, move `12` to the other side by subtracting `12` from both sides of the equation.`-6``y` `+12` `=` `6` `-6``y` `+12` `-12` `=` `6` `-12` `-6``y` `=` `-6` `12-12` cancels out Finally, remove `-6` by dividing both sides of the equation by `-6`.`-6``y` `=` `-6` `-6``y``divide-6` `=` `-6``divide-6` `y` `=` `1` `6divide6` cancels out Check our workTo confirm our answer, substitute `y=1` to the original equation.`-4(2y-3)` `=` `1/2(12-4y)` `-4[2(1)-3]` `=` `1/2[12-4(1)]` Substitute `y=1` `-4(2-3)` `=` `1/2(12-4)` `-4(-1)` `=` `1/2(8)` `4` `=` `4` Since the equation is true, the answer is correct.`y=1` -
Question 2 of 4
2. Question
Solve`2/3(n-5)=3/4(2n+10)`- `n=` (-13)
Hint
Help VideoCorrect
Correct!
Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Distributive Property
`a``(b+c)=``a``b+``a``c`Get `n` alone to the left side and all constants to the right.First, remove the fractions by finding the Least Common Denominator `(LCD)` of the denominators `4` and `3`.Multiples of `3`:`3 6 9` `12` `15`Multiples of `4`:`4 8` `12` `16 20`The `LCD` of `3` and `4` is `12`Multiply the `LCD` to both sides of the equation to remove the fractions. Use the Distributive Property.$$\frac{2}{3}(\color{#00880A}{n}-5)\color{#D800AD}{\times12}$$ `=` $$\frac{3}{4}(2\color{#00880A}{n}+10)\color{#D800AD}{\times12}$$ $$\frac{24}{3}(\color{#00880A}{n}-5)$$ `=` $$\frac{36}{4}(2\color{#00880A}{n}+10)$$ `8``(``n` `-5)` `=` `9``(2``n` `+10)` `8``n` `+``8``(-5)` `=` `9``(2``n``)+``9``(10)` `8``n` `-40` `=` `18``n` `+90` Simplify Next, move `8n` to the other side by subtracting `8n` from both sides of the equation.`8``n` `-40` `=` `18``n` `+90` `8``n` `-40` `-8n` `=` `18``n` `+90` `-8n` `-40` `=` `10``n` `+90` `8n-8n` cancels out Now, move `90` to the other side by subtracting `90` from both sides of the equation.`-40` `=` `10``n` `+90` `-40` `-90` `=` `10``n` `+90` `-90` `-130` `=` `10``n` `90-90` cancels out Finally, remove `10` by dividing both sides of the equation by `10`.`-130` `=` `10``n` `-130``divide10` `=` `10``n``divide10` `-13` `=` `n` `10divide10` cancels out `n` `=` `-13` Check our workTo confirm our answer, substitute `n=-13` to the original equation.`2/3(n-5)` `=` `3/4(2n+10)` `2/3(-13-5)` `=` `3/4(2(-13)+10)` Substitute `n=-13` `2/3(-18)` `=` `3/4(-26+10)` `(-36)/3` `=` `3/4(-16)` `-12` `=` `(-48)/4` `-12` `=` `-12` Since the equation is true, the answer is correct.`n=-13` -
Question 3 of 4
3. Question
Solve`(3(p-3))/4=(2(2p-7))/5`- `p=` (11)
Hint
Help VideoCorrect
Great Work!
Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Distributive Property
`a``(b+c)=``a``b+``a``c`Get `p` alone to the left side and all constants to the right.First, remove the fractions by finding the Least Common Denominator `(LCD)` of the denominators `4` and `5`.Multiples of `4`:`4 8 12 16` `20`Multiples of `5`:`5 10 15` `20` `25`The `LCD` of `4` and `5` is `20`Multiply the `LCD` to both sides of the equation to remove the fractions. Use the Distributive Property.$$\left[\frac{3(\color{#00880A}{p}-3)}{4}\right]\color{#D800AD}{\times20}$$ `=` $$\left[\frac{2(2\color{#00880A}{p}-7)}{5}\right]\color{#D800AD}{\times20}$$ $$\frac{20\times3}{4}(\color{#00880A}{p}-3)$$ `=` $$\frac{20\times2}{5}(2\color{#00880A}{p}-7)$$ $$\frac{60}{4}(\color{#00880A}{p}-3)$$ `=` $$\frac{40}{5}(2\color{#00880A}{p}-7)$$ `15``(``p` `-3)` `=` `8``(2``p` `-7)` `15``p` `+``15``(-3)` `=` `8``(2``p``)+``8``(-7)` `15``p` `-45` `=` `16``p` `-56` Simplify Next, move `15p` to the other side by subtracting `15p` from both sides of the equation.`15``p` `-45` `=` `16``p` `-56` `15``p` `-45` `-15p` `=` `16``p` `-56` `-15p` `-45` `=` `p` `-56` `15p-15p` cancels out Finally, move `-56` to the other side by adding `56` to both sides of the equation.`-45` `=` `p` `-56` `-45` `+56` `=` `p` `-56` `+56` `11` `=` `p` `-56+56` cancels out `p` `=` `11` Check our workTo confirm our answer, substitute `p=11` to the original equation.`(3(p-3))/4` `=` `(2(2p-7))/5` `(3(11-3))/4` `=` `(2(2(11)-7))/5` Substitute `p=11` `(3(8))/4` `=` `(2(22-7))/5` `24/4` `=` `(2(15))/5` `24/4` `=` `30/5` `6` `=` `6` Since the equation is true, the answer is correct.`p=11` -
Question 4 of 4
4. Question
Solve for `x``(2x+5)/4 = (x-1)/3`- `x=` (-19/2)
Hint
Help VideoCorrect
Well Done!
Incorrect
To solve for `x`, get `x` by itselfMultiply both sides of the equation by `4``frac{2x+5}{4}``times4` `=` `frac{x-1}{3}``times4` `frac{4(2x+5)}{4}` `=` `frac{x-1}{3}``times4` `2x+5` `=` `frac{4(x-1)}{3}` The coefficient `frac{4}{4}` cancels out `2x+5` `=` `frac{(x xx 4)-(1xx4)}{3}` Distribute the value inside the parenthesis `2x+5` `=` `frac{4x-4}{3}` Next, multiply both sides of the equation by `3``(2x+5)``times3` `=` `frac{4x-4}{3}``times3` `(2x+5)``times3` `=` `frac{3(4x-4)}{3}` `3(2x+5)` `=` `4x-4` The coefficient `frac{3}{3}` cancels out `(2x xx 3)+(5xx3)` `=` `4x-4` Distribute the value inside the parenthesis `6x+15` `=` `4x-4` Now, subtract `4x` from both sides of the equation`6x+15` `-4x` `=` `4x-4` `-4x` `6x+15` `-4x` `=` `4x``-4` `-4x` `4x-4x` cancels out `2x+15` `=` `-4` Now, subtract `15` from both sides of the equation`2x+15` `-15` `=` `-4` `-15` `2x``+15` `-15` `=` `-4` `-15` `15-15` cancels out `2x` `=` `-19` Finally, divide both sides of the equation by `2`.`2x``divide2` `=` `-19``divide2` `2``x``divide2` `=` `-19``divide2` `times2divide2` cancels out `x` `=` `-frac{19}{2}` `x=-frac{19}{2}`
Quizzes
- One Step Equations – Add and Subtract 1
- One Step Equations – Add and Subtract 2
- One Step Equations – Add and Subtract 3
- One Step Equations – Add and Subtract 4
- One Step Equations – Multiply and Divide 1
- One Step Equations – Multiply and Divide 2
- One Step Equations – Multiply and Divide 3
- One Step Equations – Multiply and Divide 4
- Two Step Equations 1
- Two Step Equations 2
- Two Step Equations 3
- Two Step Equations 4
- Multi-Step Equations 1
- Multi-Step Equations 2
- Solve Equations using the Distributive Property 1
- Solve Equations using the Distributive Property 2
- Solve Equations using the Distributive Property 3
- Equations with Variables on Both Sides 1
- Equations with Variables on Both Sides 2
- Equations with Variables on Both Sides 3
- Equations with Variables on Both Sides (Fractions) 1
- Equations with Variables on Both Sides (Fractions) 2
- Solve Equations with Variables on Both Sides using the Distributive Property 1
- Solve Equations with Variables on Both Sides using the Distributive Property 2
- Solve Equations with Variables on Both Sides using the Distributive Property 3
- Solve Equations with Variables on Both Sides using the Distributive Property 4
- Equation Word Problems 1
- Equation Word Problems 2
- Equation Word Problems 3
- Equation Word Problems 4
- Equation Word Problems (Age)
- Equation Word Problems (Money)
- Equation Word Problems (Harder)
- Equation Problems with Substitution
- Equation Problems (Geometry) 1
- Equation Problems (Geometry) 2
- Equation Problems (Perimeter)
- Equation Problems (Area)
- Change the Subject of an Equation 1
- Change the Subject of an Equation 2
- Change the Subject of an Equation 3