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

Question

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

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 2

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 3

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 4

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 5

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 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

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

(−4,−16) and (−3,−13)

Answer 7

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 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,−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 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)