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Systems of Nonlinear EquationsSystems of Nonlinear Equations
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Question 1 of 5
1. Question
Find the solution to the system of equations by graphing.
`y=4-x^2`
`y=x^2-4`
- Intersection Point `1=` (2,0, -2,0) Intersection Point `2=` (2,0, -2,0)
Correct
Great Work!
Incorrect
First set the two equations equal to each other and solve for `x``x^2-4 \ color(crimson)(+x^2)` `=` `4-x^2 \ color(crimson)(+x^2)` Add `x^2` to both sides `-4 \ color(crimson)(+x^2+x^2)` `=` `4 \ cancel(-x^2+x^2)` `2x^2-4` `=` `4` `2x^2-4 \ color(crimson)(-4)` `=` `4 \ color(crimson)(-4)` Add `-4` to both sides `2x^2-8` `=` `0` Factor out a `2` `2(x^2-4)` `=` `0` `2(x-2)(x+2)` `=` `0` Factor using the difference of squares Solve `x-2` for `x`.`x-2` `=` `0` Solve for `x` `x` `=` `2` `y` `=` `(2)^2-4` Solve for the `y` coordinate by substituting `x=2` into the second equation `y=x^2-4` `y` `=` `0` The first solution is the point `(2,0)`
Now solve `x+2` for `x`.`x+2` `=` `0` Solve for `x` `x` `=` `-2` `y` `=` `4-(-2)^2` Solve for the `y` coordinate by substituting `x=-2` into the first equation `y=4-x^2` `y` `=` `0` The second solution is the point `(-2,0)`
`:. color(darkviolet){(2,0)}` and ` color(darkviolet){(-2,0)}`
`(2,0)` and `(-2,0)` -
Question 2 of 5
2. Question
Find the solution to the system of equations by graphing.
`y=3x-2`
`y=x^2`
- Intersection Point `1=` (2,4, 1,1) Intersection Point `2=` (2,4, 1,1)
Correct
Great Work!
Incorrect
First set the two equations equal to each other and solve for `x``x^2 \ color(crimson)(-3x)` `=` `3x-2 \ color(crimson)(-3x)` Add `-3x` to both sides `x^2\ color(crimson)(-3x)` `=` `-2 \ cancel(+3x) cancel(-3x)` `x^2-3x` `=` `-2` `x^2-3x \ color(crimson)(+2)` `=` `-2 \ color(crimson)(+2)` Add `2` to both sides `x^2-3x+2` `=` `0` `(x-2)(x-1)` `=` `0` Factor Solve `x-2` for `x`.`x-2` `=` `0` Solve for `x` `x` `=` `2` `y` `=` `(2)^2` Solve for the `y` coordinate by substituting `x=2` into the second equation `y=x^2` `y` `=` `4` The first solution is the point `(2,4)`
Now solve `x-1` for `x`.`x-1` `=` `0` Solve for `x` `x` `=` `1` `y` `=` `(1)^2` Solve for the `y` coordinate by substituting `x=1` into the second equation `y=x^2` `y` `=` `1` The second solution is the point `(1,1)`
`:. color(darkviolet){(2,4)}` and ` color(darkviolet){(1,1)}`
`(2,4)` and `(1,1)` -
Question 3 of 5
3. Question
Find the solution to the system of equations by graphing.
`y=-x-6`
`y=x^2-5x-3`
- Intersection Point `1=` (3,-9, 1,-7) Intersection Point `2=` (3,-9, 1,-7)
Correct
Great Work!
Incorrect
First set the two equations equal to each other and solve for `x``x^2-5x-3 \ color(crimson)(+x)` `=` `-x-6 \ color(crimson)(+x)` Add `x` to both sides `x^2-3 \ color(crimson)(-5x+x)` `=` `-6 \ cancel(-x) cancel(+x)` `x^2-4x-3` `=` `-6` `x^2-4x-3 \ color(crimson)(+6)` `=` `4 \ color(crimson)(+6)` Add `6` to both sides `x^2-4x+3` `=` `0` `(x-3)(x-1)` `=` `0` Factor Solve `x-3` for `x`.`x-3` `=` `0` Solve for `x` `x` `=` `3` `y` `=` `-(3)-6` Solve for the `y` coordinate by substituting `x=3` into the first equation `y=-x-6` `y` `=` `-9` The first solution is the point `(3,-9)`
Now solve `x-1` for `x`.`x-1` `=` `0` Solve for `x` `x` `=` `1` `y` `=` `-(1)-6` Solve for the `y` coordinate by substituting `x=1` into the first equation `y=-x-6` `y` `=` `-7` The second solution is the point `(1,-7)`
`:. color(darkviolet){(3,-9)}` and ` color(darkviolet){(1,-7)}`
`(3,-9)` and `(1,-7)` -
Question 4 of 5
4. Question
Solve for `y`.`y=4x^2-6x``y=2x`-
`x_1 =` (0, 2)`x_2 =` (2, 0)`y_1 =` (0, 4)`y_2 =` (4, 0)
Hint
Help VideoCorrect
Well Done!
Incorrect
First, label the two equations `1` and `2` respectively.`y` `=` `4x^2-6x` Equation `1` `y` `=` `2x` Equation `2` Set Equations `1` and `2` equal to each other.`4x^2-6x` `=` `2x` `4x^2-6x``-2x` `=` `2x``-2x` Subtract `2x` from both sides `4x^2-8x` `=` `0` Simplify `4x(x-2)` `=` `0` Factor out `4x` from the left side Solve for `x``4``x` `=` `0` `x` `=` `0` `x``-2` `=` `0` `x``-2``+2` `=` `0``+2` `x` `=` `2` Now, substitute the value of `x` to solve for `y``y` `=` `2``x` `y` `=` `2``(0)` `x=0` `y` `=` `0` `y` `=` `2``x` `y` `=` `2``(2)` `x=2` `y` `=` `4` 
`x=0,2; y =0,4` -
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Question 5 of 5
5. Question
Solve for `y`.`y=6x-3``y=x^2-8x+46`-
`x =` (7)`y =` (39)
Hint
Help VideoCorrect
Exceptional!
Incorrect
First, label the two equations `1` and `2` respectively.`y` `=` `6x-3` Equation `1` `y` `=` `x^2-8x+46` Equation `2` Set Equations `1` and `2` equal to each other.`6x-3` `=` `x^2-8x+46` `6x-3``-6x+3` `=` `x^2-8x+46``-6x+3` Add `-6x+3` to both sides `0` `=` `x^2-14x+49` Simplify `x^2-14x+49` `=` `0` Since the equation is in standard form `(``a``x^2+``b``x+``c``=0)` we can factorise using the cross method.`x^2``-14``x+``49``=0`To factorise, we need to find two numbers that add to `-14` and multiply to `49`Two `-7`’s fit both conditions`-7 + (-7)` `=` `-14` `-7 xx -7` `=` `49` Read across to get the factors.`(x-7)(x-7)=0`Solve for `x`.`x``-7` `=` `0` `x``-7``+7` `=` `0``+7` `x` `=` `7` Now, substitute the value of `x` to solve for `y``y` `=` `6``x``-3` `y` `=` `6``(7)``-3` `x=7` `y` `=` `42-3` `y` `=` `39` 
`x=7; y =39` -
Quizzes
- Solve a System of Equations by Graphing
- 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