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

Question

Responses

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

no solution
no solution

infinite solutions
infinite solutions

(2−1)

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

no solution
no solution

(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

infinite solutions
infinite solutions

(−3 1/3,2/3)

Which of the following shows a system of equations equivalent to 2x+7y=14 and 4x+5y=10 ?(1 point)
Responses

2x+7y=14 and 2x+5y=5
2 x plus 7 y equals 14 and 2 x plus 5 y equals 5

4x+14y=14 and 4x+5y=10
4 x plus 14 y equals 14 and 4 x plus 5 y equals 10

2x+7y=14 and 8x+10y=20
2 x plus 7 y equals 14 and 8 x plus 10 y equals 20

4x+14y=28 and 8x+10y=10

Why is the system of equations 2x+7y=14 and 4x+5y=10 equivalent to the system of equations 4x+14y=28 and 8x+10y=20 ?(1 point)
Responses

All equations share the point (0,2).
All equations share the point left parenthesis 0 comma 2 right parenthesis .

Both equations, 2x+7y=14 and 4x+5y=10, can be multiplied by 12 to produce the equivalent system of equations.
Both equations, 2 x plus 7 y equals 14 and 4 x plus 5 y equals 10 , can be multiplied by Start Fraction 1 over 2 End Fraction to produce the equivalent system of equations.

The intercepts of the equations 2x+7y=14 and 4x+14y=28 are (0,2) and (7,0).
The intercepts of the equations 2 x plus 7 y equals 14 and 4 x plus 14 y equals 28 are left parenthesis 0 comma 2 right parenthesis and left parenthesis 7 comma 0 right parenthesis .

The intercepts of both 4x+5y=10 and 8x+10y=20 are (0,2.5) and (2,0).

Solve the following system of equations using the elimination method.

3x−4y=−4 , x+3y=−10

(1 point)
Responses

(−4,−2)
left parenthesis negative 4 comma negative 2 right parenthesis

(−16,2)
left parenthesis negative 16 comma 2 right parenthesis

(−16,−2)
left parenthesis negative 16 comma negative 2 right parenthesis

(−2,−4)

please double check and give me a answer summary

Answer 1

Let's review and summarize the solutions for each of the provided systems of equations.

1. System:

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

Solution: First, notice that the second equation can be simplified: \[ -4x - 6y = -8 \implies 4x + 6y = 8 \implies 2x + 3y = 4 \] This is the same as the first equation. Since both equations are the same, there are infinitely many solutions.
Response: infinite solutions

2. System:

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

Substitution Method: Substitute \( x \) in the second equation: \[ (y - 4) + 8y = 2 \implies 9y - 4 = 2 \implies 9y = 6 \implies y = \frac{2}{3} \] Now substitute \( y \) back to find \( x \): \[ x = \frac{2}{3} - 4 = \frac{2}{3} - \frac{12}{3} = -\frac{10}{3} \quad \left(\text{or } -3\frac{1}{3}\right) \] Response: (−3 1/3, 2/3)

3. Equivalent System:

\( 2x + 7y = 14 \) \( 4x + 5y = 10 \)

Checking Options for Equivalence: To create an equivalent system:

• Multiply the first equation by 2: \[ 4x + 14y = 28 \]
• Multiply the second equation by 2: \[ 8x + 10y = 20 \] Thus, the equivalent system is: \[ 4x + 14y = 28 \quad \text{and} \quad 8x + 10y = 20 \] Response: 4x + 14y = 28 and 8x + 10y = 20

4. Reason for Equivalence:

Reasoning: The equivalence holds because both equations represent the same geometrical lines. A scaling factor can convert the original equations to the new ones. Response: Both equations, 2x + 7y = 14 and 4x + 5y = 10, can be multiplied by 1/2 to produce the equivalent system of equations.

5. System:

\( 3x - 4y = -4 \) \( x + 3y = -10 \)

Elimination Method: Rearrange the second equation: \[ x = -10 - 3y \] Substitute \( x \) in the first equation: \[ 3(-10 - 3y) - 4y = -4 \] Simplifying: \[ -30 - 9y - 4y = -4 \implies -13y = 26 \implies y = -2 \] Substitute \( y \) back to find \( x \): \[ x = -10 - 3(-2) = -10 + 6 = -4 \] Response: (−4,−2)

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Summary of Responses:

1. infinite solutions
2. (−3 1/3, 2/3)
3. 4x + 14y = 28 and 8x + 10y = 20
4. Both equations can be multiplied by 1/2
5. (−4,−2)

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