lesson 3 rsg
two common strategies for solving systems of equations are graphing and substitution. in previous units, solving systems of linear equations has been done with these two methods. use the same methods and apply them to the systems below.
for these systems, show how to solve algebraically and with a graph. you may use graphing technology if you desire.
14. $\begin{cases} y = x^2 + 5x \\ y = 6 \end{cases}$
a. solve by substitution.
b. solve by graphing.
$\begin{cases} y = x^2 - 3x - 10 \\ y = -2x + 10 \end{cases}$
a. solve by substitution.
b. solve by graphing.

Question

lesson 3 rsg
two common strategies for solving systems of equations are graphing and substitution. in previous units, solving systems of linear equations has been done with these two methods. use the same methods and apply them to the systems below.
for these systems, show how to solve algebraically and with a graph. you may use graphing technology if you desire.
14. $\begin{cases} y = x^2 + 5x \\ y = 6 \end{cases}$
  a. solve by substitution.
  b. solve by graphing.
$\begin{cases} y = x^2 - 3x - 10 \\ y = -2x + 10 \end{cases}$
  a. solve by substitution.
  b. solve by graphing.

Accepted Answer

### Problem 14
#### Part a: Solve by substitution
# Explanation:
## Step 1: Substitute $ y = 6 $ into the first equation
We have the system $\begin{cases}y = x^2 + 5x\y = 6\end{cases}$. Substitute $ y = 6 $ into $ y=x^2 + 5x $, we get $ 6=x^2 + 5x $.
## Step 2: Rearrange into standard quadratic form
Rearrange the equation $ 6=x^2 + 5x $ to $ x^2+5x - 6=0 $.
## Step 3: Factor the quadratic equation
Factor $ x^2+5x - 6 $. We need two numbers that multiply to $- 6$ and add to $5$. The numbers are $6$ and $- 1$. So, $x^2+5x - 6=(x + 6)(x - 1)=0$.
## Step 4: Solve for $x$
Set each factor equal to zero: $x+6 = 0$ or $x - 1=0$. Solving these, we get $x=-6$ or $x = 1$.
# Answer:
The solutions are $(-6,6)$ and $(1,6)$

#### Part b: Solve by graphing
# Explanation:
## Step 1: Analyze the first equation $y=x^2 + 5x$
The equation $y=x^2+5x$ is a parabola. The vertex of a parabola $y = ax^2+bx + c$ is at $x=-\frac{b}{2a}$. Here, $a = 1$, $b = 5$, so $x=-\frac{5}{2}=-2.5$. Substitute $x=-2.5$ into $y=x^2 + 5x$, we get $y=(-2.5)^2+5\times(-2.5)=6.25-12.5=-6.25$. So the vertex is $(-2.5,-6.25)$. The parabola opens upwards since $a = 1>0$.
## Step 2: Analyze the second equation $y = 6$
The equation $y = 6$ is a horizontal line.
## Step 3: Find the intersection points
To graph, plot the parabola $y=x^2 + 5x$ and the line $y = 6$. The points where they intersect are the solutions. From the algebraic solution, we know the intersection points are $(-6,6)$ and $(1,6)$. When we graph the parabola (opening upwards, vertex at $(-2.5,-6.25)$) and the horizontal line $y = 6$, we can see that they intersect at $x=-6$ and $x = 1$ (with $y = 6$ in both cases).
# Answer:
The solutions are the points of intersection of the parabola $y=x^2 + 5x$ and the line $y = 6$, which are $(-6,6)$ and $(1,6)$

### Problem 15
#### Part a: Solve by substitution
# Explanation:
## Step 1: Substitute $y=-2x + 10$ into the first equation
We have the system $\begin{cases}y=x^2-3x - 10\y=-2x + 10\end{cases}$. Substitute $y=-2x + 10$ into $y=x^2-3x - 10$, we get $-2x + 10=x^2-3x - 10$.
## Step 2: Rearrange into standard quadratic form
Rearrange the equation $-2x + 10=x^2-3x - 10$ to $x^2-3x+2x-10 - 10=0$, which simplifies to $x^2-x - 20=0$.
## Step 3: Factor the quadratic equation
Factor $x^2-x - 20$. We need two numbers that multiply to $-20$ and add to $-1$. The numbers are $-5$ and $4$. So, $x^2-x - 20=(x - 5)(x + 4)=0$.
## Step 4: Solve for $x$
Set each factor equal to zero: $x - 5=0$ or $x + 4=0$. Solving these, we get $x = 5$ or $x=-4$.
## Step 5: Find the corresponding $y$ values
- When $x = 5$, substitute into $y=-2x + 10$, $y=-2\times5+10=0$.
- When $x=-4$, substitute into $y=-2x + 10$, $y=-2\times(-4)+10=8 + 10=18$.
# Answer:
The solutions are $(5,0)$ and $(-4,18)$

#### Part b: Solve by graphing
# Explanation:
## Step 1: Analyze the first equation $y=x^2-3x - 10$
The equation $y=x^2-3x - 10$ is a parabola. The vertex of a parabola $y=ax^2+bx + c$ is at $x =-\frac{b}{2a}$. Here, $a = 1$, $b=-3$, so $x=\frac{3}{2}=1.5$. Substitute $x = 1.5$ into $y=x^2-3x - 10$, we get $y=(1.5)^2-3\times1.5-10=2.25-4.5-10=-12.25$. So the vertex is $(1.5,-12.25)$. The parabola opens upwards since $a = 1>0$.
## Step 2: Analyze the second equation $y=-2x + 10$
The equation $y=-2x + 10$ is a line with slope $-2$ and $y$-intercept $10$.
## Step 3: Find the intersection points
To graph, plot the parabola $y=x^2-3x - 10$ and the line $y=-2x + 10$. The points where they intersect are the solutions. From the algebraic solution, we know the intersection points are $(5,0)$ and $(-4,18)$. When we graph the parabola (opening upwards, vertex at $(1.5,-12.25)$) and the line $y=-2x + 10$ (slope $-2$, $y$-intercept $10$), we can see that they intersect at $x = 5$ (with $y = 0$) and $x=-4$ (with $y = 18$).
# Answer:
The solutions are the points of intersection of the parabola $y=x^2-3x - 10$ and the line $y=-2x + 10$, which are $(5,0)$ and $(-4,18)$

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Question

Explanation:

Problem 14

Part a: Solve by substitution

Step 1: Substitute \( y = 6 \) into the first equation

Step 2: Rearrange into standard quadratic form

Step 3: Factor the quadratic equation

Step 4: Solve for \(x\)

Step 1: Analyze the first equation \(y=x^2 + 5x\)

Step 2: Analyze the second equation \(y = 6\)

Step 3: Find the intersection points

Step 1: Substitute \(y=-2x + 10\) into the first equation

Step 2: Rearrange into standard quadratic form

Step 3: Factor the quadratic equation

Step 4: Solve for \(x\)

Step 5: Find the corresponding \(y\) values

Answer:

Part b: Solve by graphing