Question

Use the substitution method to solve the system 2.5x+y=−2 , 3x+2y=0 .(1 point)
Responses

infinite solutions
infinite solutions

no solution
no solution

(−2,3)
left parenthesis negative 2 comma 3 right parenthesis

(3,−2)
left parenthesis 3 comma negative 2 right parenthesis
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Answer 1

Use the substitution method to solve the system x=y−4 , x+8y=2 .(1 point)
Responses

(2/3,−3 1/3)
left parenthesis Start Fraction 2 over 3 End Fraction comma negative 3 Start Fraction 1 over 3 End Fraction right parenthesis

no solution
no solution

infinite solutions
infinite solutions

(−3 1/3,2/3)

Answer 2

Use the substitution method to solve the system 2x+3y=4 , −4x−6y=−8 .(1 point)
Responses

infinite solutions
infinite solutions

(2−1)
left parenthesis 2 comma negative 1 right parenthesis

no solution
no solution

(−1,2)

Answer 3

Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8 , y=3x−4 ?(1 point)
Responses

(4,8)
left parenthesis 4 comma 8 right parenthesis

(3,5)
left parenthesis 3 comma 5 right parenthesis

(4,8) and (3,5)
left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis

(−4,−16) and (−3,−13)

Answer 4

Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point)
Responses

(−7,26)
left parenthesis negative 7 comma 26 right parenthesis

(5,−10) and (−7,26)
left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis

(5,−10)
left parenthesis 5 comma negative 10 right parenthesis

(5,−20) and (−7,16)
left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis

Answer 5

To solve the system of equations using the substitution method, let's start with the two equations provided:

1. \( 2.5x + y = -2 \) \quad (Equation 1)
2. \( 3x + 2y = 0 \) \quad (Equation 2)

### Step 1: Solve one of the equations for one variable

We'll solve Equation 1 for \( y \):

\[
y = -2 - 2.5x
\]

### Step 2: Substitute into the other equation

Now we substitute the expression for \( y \) into Equation 2:

\[
3x + 2(-2 - 2.5x) = 0
\]

### Step 3: Simplify and solve for \( x \)

Now, we'll simplify the equation:

\[
3x - 4 - 5x = 0
\]

Combine like terms:

\[
-2x - 4 = 0
\]

Now, isolate \( x \):

\[
-2x = 4
\]

\[
x = -2
\]

### Step 4: Substitute back to find \( y \)

Now we can substitute \( x = -2 \) back into the expression for \( y \):

\[
y = -2 - 2.5(-2)
\]

Calculate \( y \):

\[
y = -2 + 5 = 3
\]

### Step 5: Write the solution

The solution to the system of equations is \( (x, y) = (-2, 3) \).

### Final Answer

Thus, the correct response is:

\( (-2, 3) \) (left parenthesis negative 2 comma 3 right parenthesis)

Answer 6

To solve the system of equations using the substitution method, we start with the two equations provided:

1. \( x = y - 4 \) \quad (Equation 1)
2. \( x + 8y = 2 \) \quad (Equation 2)

### Step 1: Substitute Equation 1 into Equation 2

We'll substitute \( x = y - 4 \) from Equation 1 into Equation 2:

\[
(y - 4) + 8y = 2
\]

### Step 2: Simplify the equation

Now, simplify the equation:

\[
y - 4 + 8y = 2
\]

Combine like terms:

\[
9y - 4 = 2
\]

### Step 3: Solve for \( y \)

Now, isolate \( y \):

\[
9y = 2 + 4
\]

\[
9y = 6
\]

\[
y = \frac{6}{9} = \frac{2}{3}
\]

### Step 4: Substitute back to find \( x \)

Now we can substitute \( y = \frac{2}{3} \) back into Equation 1 to find \( x \):

\[
x = \frac{2}{3} - 4
\]

Convert \( 4 \) into a fraction:

\[
x = \frac{2}{3} - \frac{12}{3} = \frac{2 - 12}{3} = \frac{-10}{3}
\]

### Step 5: Write the solution

Thus, the solution to the system of equations is:

\(\left( \frac{-10}{3}, \frac{2}{3} \right)\)

Since the pairs based on the options provided do not exactly match \( \left( \frac{-10}{3}, \frac{2}{3} \right) \), I can see that \((-3 \frac{1}{3}, \frac{2}{3})\) simplifies to \(\left(-\frac{10}{3}, \frac{2}{3}\right)\).

### Final Answer

The correct response is:

\((-3 \frac{1}{3}, \frac{2}{3})\) (left parenthesis negative three and one third comma two thirds right parenthesis)

Answer 7

To solve the system of equations using the substitution method, we start with the two equations:

1. \( 2x + 3y = 4 \) \quad (Equation 1)
2. \( -4x - 6y = -8 \) \quad (Equation 2)

### Step 1: Solve one of the equations for one variable

We'll solve Equation 1 for \( y \):

\[
3y = 4 - 2x
\]

Now, divide both sides by \( 3 \):

\[
y = \frac{4 - 2x}{3}
\]

### Step 2: Substitute into the other equation

Now we substitute \( y = \frac{4 - 2x}{3} \) into Equation 2:

\[
-4x - 6\left(\frac{4 - 2x}{3}\right) = -8
\]

### Step 3: Simplify and solve for \( x \)

Now, we'll simplify the left side:

\[
-4x - \frac{24 - 12x}{3} = -8
\]

To eliminate the fraction, multiply everything by 3:

\[
-12x - (24 - 12x) = -24
\]

Distributing the negative:

\[
-12x - 24 + 12x = -24
\]

Now, combining like terms, we see:

\[
-24 = -24
\]

### Step 4: Analyze the result

The equation \( -24 = -24 \) is always true, indicating that the two equations are dependent. This means that they represent the same line. Therefore, they have **infinite solutions** along that line.

### Final Answer

Thus, the correct response is:

**infinite solutions**

Answer 8

To solve the linear-quadratic system of equations using substitution, we have:

1. \( y = x^2 - 4x + 8 \) (Equation 1)
2. \( y = 3x - 4 \) (Equation 2)

### Step 1: Substitute Equation 2 into Equation 1

Since both equations are already in the form \( y = \) ..., we can set the right sides equal to each other:

\[
3x - 4 = x^2 - 4x + 8
\]

### Step 2: Rearrange the equation

Now, rearranging this gives us:

\[
0 = x^2 - 4x + 8 - 3x + 4
\]

Combine like terms:

\[
0 = x^2 - 7x + 12
\]

### Step 3: Set the equation to standard form

Rearranging gives us:

\[
x^2 - 7x + 12 = 0
\]

### Step 4: Factor the quadratic equation

Next, we factor the polynomial:

\[
(x - 3)(x - 4) = 0
\]

### Step 5: Solve for \( x \)

Setting each factor equal to zero gives us:

1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( x - 4 = 0 \) → \( x = 4 \)

### Step 6: Find corresponding \( y \) values

Now we'll find the corresponding \( y \) values for both \( x = 3 \) and \( x = 4 \) using Equation 2.

1. For \( x = 3 \):

\[
y = 3(3) - 4 = 9 - 4 = 5
\]
Thus, one solution is \( (3, 5) \).

2. For \( x = 4 \):

\[
y = 3(4) - 4 = 12 - 4 = 8
\]
Thus, another solution is \( (4, 8) \).

### Final Solutions

The two solutions to the system are:

- \( (3, 5) \)
- \( (4, 8) \)

### Final Answer

The correct response is:

**(4, 8) and (3, 5)** (left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis)

Answer 9

To solve the linear-quadratic system using the substitution method, we have the following equations:

1. \( y = x^2 - x - 40 \) (Equation 1)
2. \( y = -3x - 5 \) (Equation 2)

### Step 1: Set the equations equal to each other

Since both equations equal \( y \), we can set the right-hand sides equal to each other:

\[
x^2 - x - 40 = -3x - 5
\]

### Step 2: Rearrange the equation

Now rearranging gives us:

\[
x^2 - x + 3x - 40 + 5 = 0
\]

Combine like terms:

\[
x^2 + 2x - 35 = 0
\]

### Step 3: Factor the quadratic equation

Next, we factor the quadratic:

\[
(x + 7)(x - 5) = 0
\]

### Step 4: Solve for \( x \)

Setting each factor equal to zero gives us:

1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

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

Now we'll find the corresponding \( y \) values for both \( x = -7 \) and \( x = 5 \) using Equation 2:

1. For \( x = -7 \):

\[
y = -3(-7) - 5 = 21 - 5 = 16
\]
Thus, one solution is \( (-7, 16) \).

2. For \( x = 5 \):

\[
y = -3(5) - 5 = -15 - 5 = -20
\]
Thus, another solution is \( (5, -20) \).

### Final Solutions

The two solutions to the system are:

- \( (5, -20) \)
- \( (-7, 16) \)

### Final Answer

The correct response is:

**(5, −20) and (−7, 16)** (left parenthesis 5 comma negative 20 right parenthesis and left parenthesis negative 7 comma 16 right parenthesis)