Question

4 - 5 additional practice
solving systems of linear and quadratic equations
solve each system of equations by graphing.

1. $y = x^{2}-3x$

$y = x + 5$

1. $y = x^{2}$

$y=-2x$

1. $y = x^{2}+6x + 1$

$y = x - 3$
solve each system of equations using elimination.

1. $y = x^{2}$

$y = x + 2$

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

$y=-x - 2$

1. $y = x^{2}-2x + 2$

$y = 2x - 2$

Explanation:

Step1: Set equations equal for problem 1

Set $x^{2}-3x=x + 5$. Rearrange to get $x^{2}-4x - 5=0$.

Step2: Factor the quadratic equation

Factor $x^{2}-4x - 5$ as $(x - 5)(x+1)=0$.

Step3: Solve for x

Set each factor equal to zero: $x - 5=0$ gives $x = 5$ and $x+1=0$ gives $x=-1$.

Step4: Find corresponding y - values

For $x = 5$, $y=x + 5=10$. For $x=-1$, $y=x + 5=4$.

For problem 2:

Step1: Set equations equal

Set $x^{2}=-2x$. Rearrange to $x^{2}+2x=0$.

Step2: Factor the quadratic equation

Factor $x^{2}+2x$ as $x(x + 2)=0$.

Step3: Solve for x

$x=0$ or $x=-2$.

Step4: Find corresponding y - values

For $x = 0$, $y = 0$. For $x=-2$, $y=-2\times(-2)=4$.

For problem 3:

Step1: Set equations equal

Set $x^{2}+6x + 1=x - 3$. Rearrange to $x^{2}+5x+4=0$.

Step2: Factor the quadratic equation

Factor $x^{2}+5x + 4$ as $(x + 1)(x + 4)=0$.

Step3: Solve for x

$x=-1$ or $x=-4$.

Step4: Find corresponding y - values

For $x=-1$, $y=-1 - 3=-4$. For $x=-4$, $y=-4 - 3=-7$.

For problem 4:

Step1: Set equations equal

Set $x^{2}=x + 2$. Rearrange to $x^{2}-x - 2=0$.

Step2: Factor the quadratic equation

Factor $x^{2}-x - 2$ as $(x - 2)(x+1)=0$.

Step3: Solve for x

$x = 2$ or $x=-1$.

Step4: Find corresponding y - values

For $x = 2$, $y=2 + 2=4$. For $x=-1$, $y=-1 + 2=1$.

For problem 5:

Step1: Set equations equal

Set $x^{2}-4=-x - 2$. Rearrange to $x^{2}+x - 2=0$.

Step2: Factor the quadratic equation

Factor $x^{2}+x - 2$ as $(x + 2)(x - 1)=0$.

Step3: Solve for x

$x=-2$ or $x = 1$.

Step4: Find corresponding y - values

For $x=-2$, $y=-(-2)-2=0$. For $x = 1$, $y=-1 - 2=-3$.

For problem 6:

Step1: Set equations equal

Set $x^{2}-2x + 2=2x-2$. Rearrange to $x^{2}-4x + 4=0$.

Step2: Factor the quadratic equation

Factor $x^{2}-4x + 4$ as $(x - 2)^{2}=0$.

Step3: Solve for x

$x = 2$.

Step4: Find corresponding y - values

$y=2\times2-2=2$.

Answer:

Problem 1: $(-1,4),(5,10)$
Problem 2: $(0,0),(-2,4)$
Problem 3: $(-1,-4),(-4,-7)$
Problem 4: $(-1,1),(2,4)$
Problem 5: $(-2,0),(1,-3)$
Problem 6: $(2,2)$