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Equation Problems with Substitution 1Equation Problems with Substitution 1
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Question 1 of 6
1. Question
Find `w` if the perimeter of the rectangle below is `22` cm.
- `w=` (4)
Hint
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Perimeter of a Rectangle
`P=2``L` `+2``W`Form an equation using the formula for the Perimeter of a Rectangle.
`P=22`cm`L=7`cm`w` cm`P` `=` `2``L` `+2``W` `22` `=` `2(``7``)+2``w` Substitute the values `22` `=` `14+2w` Simplify To solve for `w`, it needs to be alone on one side.Start by moving `14` to the other side by subtracting `14` from both sides of the equation.`22` `=` `14+2w` `22` `-14` `=` `14+2w` `-14` `8` `=` `2``w` `14-14` cancels out Finally, remove `2` by dividing both sides of the equation by `2`.`8` `=` `2``w` `8``divide2` `=` `2``w``divide2` `4` `=` `w` `w` `=` `4` Check our workTo confirm our answer, substitute `w=4` to the formed equation.`22` `=` `14+2w` `22` `=` `14+2(4)` Substitute `w=4` `22` `=` `14+8` `22` `=` `22` Since the equation is true, the answer is correct.`w=4` -
Question 2 of 6
2. Question
Find the height `(h)` of the triangle below if its area is `45` cm².
- (9) cm
Hint
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Area of a Triangle
`A=1/2``b``h`First, label the values and form an equation using the Area of a Triangle formula.
`b=10` cm`h=?`cm`A=45`cm²`A` `=` `1/2``b``h` `45` `=` `1/2``10``h` Substitute the values `45` `=` `5h` Simplify To solve for `h`, it needs to be alone on one side.Remove `5` by dividing both sides of the equation by `5`.`45` `=` `5``h` `45``divide5` `=` `5``h``divide5` `9` `=` `h` `5-:5` cancels out `h` `=` `9` cm `9` cm -
Question 3 of 6
3. Question
Find the height `(h)` of the trapezium below if `a=7`, `b=9` and its area `(A)` is `72`.
- `h=` (9)
Hint
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Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Area of a Trapezium
`A=1/2``h``(``a` `+``b``)`First, label the values and form an equation using the Area of a Trapezium formula.
`A=72``a=7``b=9``h=?``A` `=` `1/2``h``(``a` `+``b``)` `72` `=` `1/2``h``(``7` `+``9``)` Substitute the values `72` `=` `1/2 h(16)` Simplify `72` `=` `8h` To solve for `h`, it needs to be alone on one side.Remove `8` by dividing both sides of the equation by `8`.`72` `=` `8``h` `72``divide8` `=` `8``h``divide8` `9` `=` `h` `8-:8` cancels out `h` `=` `9` `h=9` -
Question 4 of 6
4. Question
Given that `V=u+at`, find `a` using the following values:`V=178``u=28``t=10`- `a=` (15)
Hint
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Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.First, list the values and form an equation using the given formula for `V`.`V=178``u=28``t=10``V` `=` `u` `+a``t` `178` `=` `28` `+a``(10)` Substitute the values `178` `=` `28+10a` Simplify To solve for `a`, it needs to be alone on one side.Start by moving `28` to the other side by subtracting `28` from both sides of the equation.`178` `=` `28+10``a` `178` `-28` `=` `28+10``a` `-28` `150` `=` `10``a` `28-28` cancels out Finally, remove `10` by dividing both sides of the equation by `10`.`150` `=` `10``a` `150``divide10` `=` `10``a``divide10` `15` `=` `a` `10divide10` cancels out `a` `=` `15` `a=15` -
Question 5 of 6
5. Question
Given that `A=(x+y)/2`, find `y` using the following values:`A=51``x=67`- `y=` (35)
Hint
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Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.First, list the values and form an equation using the given formula for `A`.`A=51``x=67``A` `=` $$\frac{\color{#007DDC}{x}+y}{2}$$ `51` `=` $$\frac{\color{#007DDC}{67}+y}{2}$$ Substitute the values To solve for `y`, it needs to be alone on one side.Start by removing `1/2` by multiplying both sides of the equation by `2`.`51` `=` $$\frac{67+\color{#00880A}{y}}{2}$$ `51``times2` `=` $$\left(\frac{67+\color{#00880A}{y}}{2}\right)\color{#CC0000}{\times2}$$ `102` `=` `67+``y` `1/2times2` cancels out Finally, move `67` to the other side by subtracting `67` from both sides of the equation.`102` `=` `67+``y` `102` `-67` `=` `67+``y` `-67` `35` `=` `y` `67-67` cancels out `y` `=` `35` Check our workTo confirm our answer, substitute `y=35` to the original equation.`51` `=` `(67+y)/2` `51` `=` `(67+35)/2` Substitute `y=35` `51` `=` `102/2` `51` `=` `51` Since the equation is true, the answer is correct.`y=35` -
Question 6 of 6
6. Question
Convert `30°C` to `F` using the formula below:`C=5/9 (F-32)`- (86)`°F`
Hint
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Excellent!
Incorrect
Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.First, use the formula and the given value to form an equation.`C=30°``C` `=` `5/9 (F-32)` `30` `=` `5/9 (F-32)` Substitute `C` To solve for `F`, it needs to be alone on one side.Start by removing `1/9` by multiplying both sides of the equation by `9`.`30` `=` `5/9 (``F` `-32)` `30``times9` `=` `5/9 (``F` `-32)``times9` `270` `=` `5(``F` `-32)` `1/9times9` cancels out Next, remove `5` by dividing both sides of the equation by `5`.`270` `=` `5(``F` `-32)` `270``divide5` `=` `5(``F` `-32)``divide5` `54` `=` `F` `-32` `5divide5` cancels out Finally, move `32` to the other side by adding `32` to both sides of the equation.`54` `=` `F` `-32` `54` `+32` `=` `F` `-32` `+32` `86` `=` `F` `-32+32` cancels out `F` `=` `86°` Therefore, `30°C` is equal to `86°F``86°F`
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 – Variables on Both Sides (Distributive Property) 1
- Solve Equations – Variables on Both Sides (Distributive Property) 2
- Solve Equations – Variables on Both Sides (Distributive Property) 3
- Solve Equations – Variables on Both Sides (Distributive Property) 4
- Writing Equations 1
- Writing Equations 2
- Writing Equations 3
- Writing Equations 4
- Equation Word Problems (Age) 1
- Equation Word Problems (Money) 1
- Equation Word Problems (Harder) 1
- Equation Problems with Substitution 1
- Equation Problems (Geometry) 1
- Equation Problems (Geometry) 2
- Equation Problems (Perimeter)
- Equation Problems (Area)
- Solve for a Variable or Formula 1
- Solve for a Variable or Formula 2
- Solve for a Variable or Formula 3