Substitution Method 3
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Question 1 of 4
1. Question
Solve the following systems of equations by substitution.`8a+5b=6``4a+2b=4`-
`a=` (2)`b=` (-2)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Substitution Method
- `1)` make one variable the subject
- `2)` substitute into second equation
- `3)` solve the second equation
- `4)` substitute back
First, label the two equations `1` and `2` respectively.`8a+5b` `=` `6` Equation `1` `4a+2b` `=` `4` Equation `2` Next, solve for `b` in Equation `2`.`4a+2b` `=` `4` `(4a``div2``)+(2b``div2``)` `=` `4``div2` Divide the values of both sides by `2` `2a+b` `-2a` `=` `2` `-2a` Subtract `2a` from both sides `b` `=` `2-2a` Simplify Substitute `b` into Equation `1`.`8a+5``b` `=` `6` Equation `1` `8a+5``(2-2a)` `=` `6` `b=2-2a` `8a+10-10a` `=` `6` Distribute `5` inside the parenthesis `10-2a` `=` `6` Simplify `10-2a` `-10` `=` `6` `-10` Solve for `a` `-2a` `=` `-4` `-2a` `div(-2)` `=` `-4` `div(-2)` Divide both sides by `-2` `a` `=` `2` Now, substitute the value of `a` into Equation `2``4``a` `+2b` `=` `4` Equation `1` `4``(2)` `+2b` `=` `4` `a=2` `8+2b` `-8` `=` `4` `-8` Subtract `8` from both sides `2b``div2` `=` `-4``div2` Divide both sides by `2` `b` `=` `-2` `a=2, b=-2` -
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Question 2 of 4
2. Question
Solve the following systems of equations by substitution.`2x+5y=17``3x+3y=12`-
`x=` (1)`y=` (3)
Hint
Help VideoCorrect
Excellent!
Incorrect
Substitution Method
- `1)` make one variable the subject
- `2)` substitute into second equation
- `3)` solve the second equation
- `4)` substitute back
First, label the two equations `1` and `2` respectively.`2x+5y` `=` `17` Equation `1` `3x+3y` `=` `12` Equation `2` Next, solve for `x` in Equation `2`.`3x+3y` `=` `12` `(3x``div3``)+(3y``div3``)` `=` `12``div3` Divide the values of both sides by `3` `x+y` `-y` `=` `4` `-y` Subtract `y` from both sides `x` `=` `4-y` Simplify Substitute `x` into Equation `1`.`2``x` `+5y` `=` `17` Equation `1` `2``(4-y)` `+5y` `=` `17` `x=4-y` `8-2y+5y` `=` `17` Distribute `2` inside the parenthesis `8+3y` `=` `17` Simplify `8+3y` `-8` `=` `17` `-8` Solve for `y` `3y` `=` `9` `3y` `div3` `=` `9` `div3` Divide both sides by `3` `y` `=` `3` Now, substitute the value of `y` into Equation `2``3x+3``y` `=` `12` Equation `2` `3x+3``(3)` `=` `12` `y=3` `3x+9` `-9` `=` `12` `-9` Subtract `9` to both sides `3x` `div3` `=` `3` `div3` Divide both sides by `3` `x` `=` `1` `x=1, y=3` -
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Question 3 of 4
3. Question
Solve the following systems of equations by substitution.`3x+7y=4``2x+3y=1`-
`x=` (-1)`y=` (1)
Hint
Help VideoCorrect
Correct!
Incorrect
Substitution Method
- `1)` make one variable the subject
- `2)` substitute into second equation
- `3)` solve the second equation
- `4)` substitute back
First, label the two equations `1` and `2` respectively.`3x+7y` `=` `4` Equation `1` `2x+3y` `=` `1` Equation `2` Next, solve for `x` in Equation `2`.`2x+3y` `=` `1` `2x+3y` `-3y` `=` `1``-3y` Subtract `3y` from both sides `2x` `div2` `=` `(1-3y)` `div2` Divide both sides by `2` `x` `=` `(1-3y)/2` Simplify Substitute `x` into Equation `1`.`3``x` `+7y` `=` `4` Equation `1` `3``((1-3y)/2)` `+7y` `=` `4` `x=(1-3y)/2` $$\left(3\left(\frac{1-3y}{2}\right)\color{#CC0000}{\times2}\right)+(7y\color{#CC0000}{\times2})$$ `=` `4``times2` Multiply all values by `2` `3(1-3y)+14y` `=` `8` `3-9y+14y` `=` `8` Distribute `3` inside the parenthesis `3+5y` `=` `8` Simplify `3+5y` `-3` `=` `8` `-3` Solve for `y` `5y` `=` `5` `5y` `div5` `=` `5` `div5` Divide both sides by `5` `y` `=` `1` Now, substitute the value of `y` into Equation `2``2x+3``y` `=` `1` Equation `2` `2x+3``(1)` `=` `1` `y=1` `2x+3` `-3` `=` `1` `-3` Subtract `3` from both sides `2x` `div2` `=` `-2` `div2` Divide both sides by `2` `x` `=` `-1` `x=-1, y=1` -
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Question 4 of 4
4. Question
Solve the following systems of equations by substitution.`5a+2b=4``2a-3b=13`-
`a=` (2)`b=` (-3)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Substitution Method
- `1)` make one variable the subject
- `2)` substitute into second equation
- `3)` solve the second equation
- `4)` substitute back
First, label the two equations `1` and `2` respectively.`5a+2b` `=` `4` Equation `1` `2a-3b` `=` `13` Equation `2` Next, solve for `a` in Equation `2`.`2a-3b` `=` `13` `2a-3b` `+3b` `=` `13``+3b` Add `3b` to both sides `2a` `div2` `=` `(13+3b)` `div2` Divide both sides by `2` `a` `=` `(13+3b)/2` Simplify Substitute `a` into Equation `1`.`5``a` `+2b` `=` `4` Equation `1` `5``((13+3b)/2)` `+2b` `=` `4` `a=(13+3b)/2` $$\left(5\left(\frac{13+3b}{2}\right)\color{#CC0000}{\times2}\right)+(2b\color{#CC0000}{\times2})$$ `=` `4``times2` Multiply all values by `2` `5(13+3b)+4b` `=` `8` `65+15b+4b` `=` `8` Distribute `5` inside the parenthesis `65+19b` `=` `8` Simplify `65+19b` `-65` `=` `8` `-65` Solve for `b` `19b` `=` `-57` `19b` `div19` `=` `-57` `div19` Divide both sides by `19` `b` `=` `-3` Now, substitute the value of `b` into Equation `2``2a-3``b` `=` `13` Equation `2` `2a-3``(-3)` `=` `13` `b=-3` `2a+9` `-9` `=` `13` `-9` Subtract `9` from both sides `2a` `div2` `=` `4` `div2` Divide both sides by `2` `a` `=` `2` `a=2, b=-3` -
Quizzes
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- Substitution Method 1
- Substitution Method 2
- Substitution Method 3
- Substitution Method 4
- Elimination Method 1
- Elimination Method 2
- Elimination Method 3
- Elimination Method 4
- Systems of Nonlinear Equations
- Systems of Equations Word Problems 1
- Systems of Equations Word Problems 2
- 3 Variable Systems of Equations – Substitution Method
- 3 Variable Systems of Equations – Elimination Method