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

Question

Responses

infinite solutions
infinite solutions

no solution
no solution

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

(−1,2)

Answer 1

To solve the system of equations using the substitution method, we first write both equations:

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

To check for consistency, we can manipulate the second equation. Notice that the second equation can be rewritten:

\[ -4x - 6y = -8 \]

We can divide the entire equation by -2: \[ 2x + 3y = 4 \]

Now, we see that both equations are actually the same:

\[ 2x + 3y = 4 \] \[ 2x + 3y = 4 \]

Since both equations represent the same line, this means that there are infinitely many solutions for this system of equations. Any point \( (x, y) \) that satisfies the first equation will also satisfy the second.

Thus, the solution to the system is:

Infinite solutions

Answer 2

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

no solution
no solution

(23,−313)
left parenthesis Start Fraction 2 over 3 End Fraction comma negative 3 Start Fraction 1 over 3 End Fraction right parenthesis

infinite solutions
infinite solutions

(−313,23)
left parenthesis negative 3 Start Fraction 1 over 3 End Fraction comma Start Fraction 2 over 3 End Fraction right parenthesis

Answer 3

To solve the system of equations using the substitution method, we first write down the equations:

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

Next, we can substitute the expression for \( x \) from the first equation into the second equation:

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

Now combine like terms:

\[ y - 4 + 8y = 2 \] \[ 9y - 4 = 2 \]

Next, isolate \( y \) by adding 4 to both sides:

\[ 9y = 6 \]

Now, divide by 9:

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

Now that we have \( y \), we can substitute back into the first equation to find \( x \):

\[ x = y - 4 \] \[ x = \frac{2}{3} - 4 \] \[ x = \frac{2}{3} - \frac{12}{3} = \frac{2 - 12}{3} = \frac{-10}{3} \]

Thus, we have:

\[ x = -\frac{10}{3}, \quad y = \frac{2}{3} \]

So the solution to the system is:

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

Based on the provided options, the correct response is:

(−3⅓, 2⅓) (which is equivalent to \(\left(-\frac{10}{3}, \frac{2}{3}\right)\)).

Answer 4

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

infinite solutions
infinite solutions

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

no solution
no solution

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

Answer 5

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

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

Step 1: Solve one of the equations for one variable

Let's solve the first equation for \( y \):

\[ y = -2 - 2.5x \]

Step 2: Substitute into the second equation

Now we substitute this expression for \( y \) into the second equation:

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

Distributing the \( 2 \):

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

Combine like terms:

\[ -2x - 4 = 0 \]

Step 3: Solve for \( x \)

Add \( 4 \) to both sides:

\[ -2x = 4 \]

Now divide by \(-2\):

\[ x = -2 \]

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

Now that we have \( x \), we substitute it back into the equation we derived for \( y \):

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

Calculate:

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

Step 5: Write the solution

The solution to the system of equations is:

\[ (x, y) = (-2, 3) \]

Based on the provided options, the answer is:

(−2, 3).

Answer 6

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,−16) and (−3,−13)
left parenthesis negative 4 comma negative 16 right parenthesis and left parenthesis negative 3 comma negative 13 right parenthesis

(4,8) and (3,5)

Answer 7

To solve the linear-quadratic system using the substitution method, we start with the two equations:

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

Step 1: Set the equations equal to each other

Since both equations equal \( y \), we can set them equal to each other:

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

Step 2: Rearrange the equation

Rearranging gives us:

\[ x^2 - 4x - 3x + 8 + 4 = 0 \] \[ x^2 - 7x + 12 = 0 \]

Step 3: Factor the quadratic equation

We can factor the quadratic:

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

Step 4: Solve for \( x \)

Setting each factor to zero gives us the solutions for \( x \):

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

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

Now we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We can use either equation, but let's use \( y = 3x - 4 \):

1. For \( x = 3 \): \[ y = 3(3) - 4 = 9 - 4 = 5 \] So one solution is \( (3, 5) \).

2. For \( x = 4 \): \[ y = 3(4) - 4 = 12 - 4 = 8 \] So the other solution is \( (4, 8) \).

Step 6: Write the solutions

The solutions to the system are:

• \( (3, 5) \)
• \( (4, 8) \)

Based on the given options, the correct response is:

(4, 8) and (3, 5).

Answer 8

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

Answer 9

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

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

Step 1: Set the equations equal to each other

Since both equations equal \( y \), we can set them equal to each other:

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

Step 2: Rearrange the equation

Rearranging gives us:

\[ x^2 - x + 3x - 40 + 5 = 0 \] \[ x^2 + 2x - 35 = 0 \]

Step 3: Factor the quadratic equation

Next, we can factor the quadratic:

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

Step 4: Solve for \( x \)

Setting each factor to zero gives us the solutions for \( x \):

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

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

Now we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. Let's use \( y = -3x - 5 \):

1. For \( x = -7 \): \[ y = -3(-7) - 5 = 21 - 5 = 16 \] So one solution is \( (-7, 16) \).

2. For \( x = 5 \): \[ y = -3(5) - 5 = -15 - 5 = -20 \] So the other solution is \( (5, -20) \).

Step 6: Write the solutions

The solutions to the system are:

• \( (5, -20) \)
• \( (-7, 16) \)

Based on the given options, the correct response is:

(5,−20) and (−7,16).