Systems of Nonlinear Equations

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Question 1 of 5

Find the solution to the system of equations by graphing.

$y = 4 - x^{2}$ $y=4-x^2$

$y = x^{2} - 4$ $y=x^2-4$

Intersection Point $1 =$ $1=$ (2,0, -2,0) Intersection Point $2 =$ $2=$ (2,0, -2,0)

Question 2 of 5

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)

Question 3 of 5

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)

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)
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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 $y$

$y$ $=$ $2$ $x$

$y$ $=$ $2$ $(0)$ $x=0$

$y$ $=$ $0$

$y$ $=$ $2$ $x$

$y$ $=$ $2$ $(2)$ $x=2$

$y$ $=$ $4$

$x=0,2; y =0,4$
Question 5 of 5

5. Question
Solve for $y$ .

$y=6x-3$

$y=x^2-8x+46$
- $x =$ (7)
  
  $y =$ (39)
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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 $y$

$y$ $=$ $6$ $x$ $-3$

$y$ $=$ $6$ $(7)$ $-3$ $x=7$

$y$ $=$ $42-3$

$y$ $=$ $39$

$x=7; y =39$

$x^{2} - 4 + x^{2}$ $x^2-4 \ color(crimson)(+x^2)$	$=$ $=$	$4 - x^{2} + x^{2}$ $4-x^2 \ color(crimson)(+x^2)$	Add $x^{2}$ $x^2$ to both sides
$- 4 + x^{2} + x^{2}$ $-4 \ color(crimson)(+x^2+x^2)$	$=$ $=$	$4 \ cancel(-x^2+x^2)$
$2x^2-4$	$=$	$4$
$2x^2-4 \ color(crimson)(-4)$	$=$	$4 \ color(crimson)(-4)$	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

$x^2 \ color(crimson)(-3x)$	$=$	$3x-2 \ color(crimson)(-3x)$	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

$x^2-5x-3 \ color(crimson)(+x)$	$=$	$-x-6 \ color(crimson)(+x)$	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

Systems of Nonlinear Equations

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Quiz summary

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1. Question

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2. Question

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3. Question

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4. Question

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5. Question

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Quizzes

$x-2$	$=$	$0$	Solve for $x$
$x$	$=$	$2$
$y$	$=$	$(2)^2-4$	Solve for the $y$ coordinate by substituting $x=2$ into the second equation $y=x^2-4$
$y$	$=$	$0$

$x+2$	$=$	$0$	Solve for $x$
$x$	$=$	$-2$
$y$	$=$	$4-(-2)^2$	Solve for the $y$ coordinate by substituting $x=-2$ into the first equation $y=4-x^2$
$y$	$=$	$0$

$x-2$	$=$	$0$	Solve for $x$
$x$	$=$	$2$
$y$	$=$	$(2)^2$	Solve for the $y$ coordinate by substituting $x=2$ into the second equation $y=x^2$
$y$	$=$	$4$

$x-1$	$=$	$0$	Solve for $x$
$x$	$=$	$1$
$y$	$=$	$(1)^2$	Solve for the $y$ coordinate by substituting $x=1$ into the second equation $y=x^2$
$y$	$=$	$1$

$x-3$	$=$	$0$	Solve for $x$
$x$	$=$	$3$
$y$	$=$	$-(3)-6$	Solve for the $y$ coordinate by substituting $x=3$ into the first equation $y=-x-6$
$y$	$=$	$-9$

$x-1$	$=$	$0$	Solve for $x$
$x$	$=$	$1$
$y$	$=$	$-(1)-6$	Solve for the $y$ coordinate by substituting $x=1$ into the first equation $y=-x-6$
$y$	$=$	$-7$

$4x^2-6x$	$=$	$2x$
$4x^2-6x$ $-2x$	$=$	$2x$ $-2x$	Subtract $2x$ from both sides
$4x^2-8x$	$=$	$0$	Simplify
$4x(x-2)$	$=$	$0$	Factor out $4x$ from the left side

$6x-3$	$=$	$x^2-8x+46$
$6x-3$ $-6x+3$	$=$	$x^2-8x+46$ $-6x+3$	Add $-6x+3$ to both sides
$0$	$=$	$x^2-14x+49$	Simplify
$x^2-14x+49$	$=$	$0$

$y$	$=$	$6$ $x$ $-3$
$y$	$=$	$6$ $(7)$ $-3$	$x=7$
$y$	$=$	$42-3$
$y$	$=$	$39$

$-7 + (-7)$	$=$	$-14$
$-7 xx -7$	$=$	$49$