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Equation Problems with Substitution 1Equation Problems with Substitution 1
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Question 1 of 6
1. Question
Find ww if the perimeter of the rectangle below is 2222 cm.- w=w= (4)
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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=2P=2LL +2+2WWForm an equation using the formula for the Perimeter of a Rectangle.P=22P=22cmL=7L=7cmww cmPP == 22LL +2+2WW 2222 == 2(2(77)+2)+2ww Substitute the values 2222 == 14+2w14+2w Simplify To solve for ww, it needs to be alone on one side.Start by moving 1414 to the other side by subtracting 1414 from both sides of the equation.2222 == 14+2w14+2w 2222 -14−14 == 14+2w14+2w -14−14 88 == 22ww 14-1414−14 cancels out Finally, remove 22 by dividing both sides of the equation by 22.88 == 22ww 88÷2÷2 == 22ww÷2÷2 44 == ww ww == 44 Check our workTo confirm our answer, substitute w=4w=4 to the formed equation.2222 == 14+2w14+2w 2222 == 14+2(4)14+2(4) Substitute w=4w=4 2222 == 14+814+8 2222 == 2222 Since the equation is true, the answer is correct.w=4w=4 -
Question 2 of 6
2. Question
Find the height (h)(h) of the triangle below if its area is 4545 cm².- (9) cm
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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=12A=12bbhhFirst, label the values and form an equation using the Area of a Triangle formula.b=10b=10 cmh=?h=?cmA=45A=45cm²AA == 1212bbhh 4545 == 12121010hh Substitute the values 4545 == 5h5h Simplify To solve for hh, it needs to be alone on one side.Remove 55 by dividing both sides of the equation by 55.4545 == 55hh 4545÷5÷5 == 55hh÷5÷5 99 == hh 5÷55÷5 cancels out hh == 99 cm 99 cm -
Question 3 of 6
3. Question
Find the height (h)(h) of the trapezium below if a=7a=7, b=9b=9 and its area (A)(A) is 7272.- h=h= (9)
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.Area of a Trapezium
A=12A=12hh((aa ++bb))First, label the values and form an equation using the Area of a Trapezium formula.A=72A=72a=7a=7b=9b=9h=?h=?AA == 1212hh((aa ++bb)) 7272 == 1212hh((77 ++99)) Substitute the values 7272 == 12h(16)12h(16) Simplify 7272 == 8h8h To solve for hh, it needs to be alone on one side.Remove 88 by dividing both sides of the equation by 88.7272 == 88hh 7272÷8÷8 == 88hh÷8÷8 99 == hh 8÷88÷8 cancels out hh == 99 h=9h=9 -
Question 4 of 6
4. Question
Given that V=u+atV=u+at, find aa using the following values:V=178V=178u=28u=28t=10t=10- a=a= (15)
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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 VV.V=178V=178u=28u=28t=10t=10VV == uu +a+att 178178 == 2828 +a+a(10)(10) Substitute the values 178178 == 28+10a28+10a Simplify To solve for aa, it needs to be alone on one side.Start by moving 2828 to the other side by subtracting 2828 from both sides of the equation.178178 == 28+1028+10aa 178178 -28−28 == 28+1028+10aa -28−28 150150 == 1010aa 28-2828−28 cancels out Finally, remove 1010 by dividing both sides of the equation by 1010.150150 == 1010aa 150150÷10÷10 == 1010aa÷10÷10 1515 == aa 10÷1010÷10 cancels out aa == 1515 a=15a=15 -
Question 5 of 6
5. Question
Given that A=x+y2A=x+y2, find yy using the following values:A=51A=51x=67x=67- y=y= (35)
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Chapters- Chapters
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 AA.A=51A=51x=67x=67AA == x+y2x+y2 5151 == 67+y267+y2 Substitute the values To solve for yy, it needs to be alone on one side.Start by removing 1212 by multiplying both sides of the equation by 22.5151 == 67+y267+y2 5151×2×2 == (67+y2)×2(67+y2)×2 102102 == 67+67+yy 12×212×2 cancels out Finally, move 6767 to the other side by subtracting 6767 from both sides of the equation.102102 == 67+67+yy 102102 -67−67 == 67+67+yy -67−67 3535 == yy 67-6767−67 cancels out yy == 3535 Check our workTo confirm our answer, substitute y=35y=35 to the original equation.5151 == 67+y267+y2 5151 == 67+35267+352 Substitute y=35y=35 5151 == 10221022 5151 == 5151 Since the equation is true, the answer is correct.y=35y=35 -
Question 6 of 6
6. Question
Convert 30°C30°C to FF using the formula below:C=59(F-32)C=59(F−32)- (86)°F°F
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Chapters- Chapters
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=30°CC == 59(F-32)59(F−32) 3030 == 59(F-32)59(F−32) Substitute CC To solve for FF, it needs to be alone on one side.Start by removing 1919 by multiplying both sides of the equation by 99.3030 == 59(59(FF -32)−32) 3030×9×9 == 59(59(FF -32)−32)×9×9 270270 == 5(F -32) 19×9 cancels out Next, remove 5 by dividing both sides of the equation by 5.270 = 5(F -32) 270÷5 = 5(F -32)÷5 54 = F -32 5÷5 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°F86°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