What operation needs to be performed to go from the equation 2x+7y=14

Question

to the equivalent equation 4x+14y=28
.(1 point)
The equation 2x+7y=14
should be multiplied by
to produce the equivalent equation 4x+14y=28
.

Answer 1

To go from the equation \( 2x + 7y = 14 \) to the equivalent equation \( 4x + 14y = 28 \), you need to multiply the entire equation by \( 2 \).

So, the statement would be:

The equation \( 2x + 7y = 14 \) should be multiplied by \( 2 \) to produce the equivalent equation \( 4x + 14y = 28 \).

Answer 2

Solve the following system of equations using the elimination method.

2x−3y=14
, 4x+3y=10

(1 point)

Answer 3

To solve the system of equations using the elimination method, we have the following two equations:

1. \( 2x - 3y = 14 \) \quad (Equation 1)
2. \( 4x + 3y = 10 \) \quad (Equation 2)

First, let's manipulate these equations to eliminate one of the variables. Since the coefficients of \( y \) in the two equations have opposite signs, we can add the two equations directly after multiplying Equation 1 by \( 1 \) (to keep it as it is) and Equation 2 by \( 1 \) (to keep it as it is).

Step 1: Prepare to eliminate \( y \)

We can add the two equations together:

\[ (2x - 3y) + (4x + 3y) = 14 + 10 \]

Step 2: Combine the equations

This results in:

\[ 2x + 4x - 3y + 3y = 14 + 10 \] \[ 6x = 24 \]

Step 3: Solve for \( x \)

Now, divide both sides by \( 6 \):

\[ x = \frac{24}{6} = 4 \]

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

Now that we have \( x = 4 \), substitute this value back into either of the original equations. We'll use Equation 1:

\[ 2(4) - 3y = 14 \]

This simplifies to:

\[ 8 - 3y = 14 \]

Step 5: Solve for \( y \)

Subtract \( 8 \) from both sides:

\[ -3y = 14 - 8 \] \[ -3y = 6 \]

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

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

Final Solution

The solution to the system of equations is:

\[ (x, y) = (4, -2) \]

Answer 4

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

Answer 5

1. Substitution Method for the First System of Equations

Given System:

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

Let's simplify the second equation: \[ -4x - 6y = -8 \implies 4x + 6y = 8 \quad (\text{just multiply by } -1) \] We can see that the second equation is just double the first equation: \[ 4x + 6y = 8 \text{ can be obtained by multiplying } (2x + 3y = 4) \text{ by } 2. \] This means the two equations represent the same line, hence there are infinite solutions.

• Response: infinite solutions

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2. Substitution Method for the Second System of Equations

Given System:

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

Substituting \( x \) from the first equation into the second: \[ (y - 4) + 8y = 2 \] Combine like terms: \[ y - 4 + 8y = 2 \implies 9y - 4 = 2 \] Add \( 4 \) to both sides: \[ 9y = 6 \implies y = \frac{6}{9} = \frac{2}{3} \] Now substitute back to find \( x \): \[ x = y - 4 = \frac{2}{3} - 4 = \frac{2}{3} - \frac{12}{3} = \frac{-10}{3} \] So the answer to the second system is:

• Response: (2/3, −3 1/3)

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3. Equivalent System of Equations

Given Equations:

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

To check for equivalence, let's identify each option:

1. \( 2x + 7y = 14 \) and \( 2x + 5y = 5 \) - Not equivalent.
2. \( 4x + 14y = 14 \) and \( 4x + 5y = 10 \) - Not equivalent.
3. \( 2x + 7y = 14 \) and \( 8x + 10y = 20 \) - Not equivalent (not properly scaled).
4. \( 4x + 14y = 28 \) and \( 8x + 10y = 20 \) - Correct, because you could arrive at these by multiplying.

So the correct response would be:

• Response: \( 4x + 14y = 28 \) and \( 8x + 10y = 20 \)

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4. Why is the system equivalent?

The first system is equivalent to the second system because they represent the same relationships. The simplest option is:

• Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( 2 \) to produce another equivalent form.
• Response: Both equations, \( 2x + 7y = 14 \) and \( 4x + 5y = 10 \), can be multiplied by \( \frac{1}{2} \) to produce the equivalent system of equations.

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5. Elimination Method for the Last System of Equations

Given System:

1. \( 3x - 4y = -4 \)
2. \( x + 3y = -10 \)

Let's eliminate \( x \) by manipulating the second equation:

First, rearranging the second equation: \[ x = -10 - 3y \]

Now substitute \( x \) in the first equation: \[ 3(-10 - 3y) - 4y = -4 \] Distribute \( 3 \): \[ -30 - 9y - 4y = -4 \] Combine like terms: \[ -30 - 13y = -4 \] Add \( 30 \) to both sides: \[ -13y = 26 \] So: \[ y = -2 \]

Now substitute back to find \( x \): \[ x = -10 - 3(-2) = -10 + 6 = -4 \]

So the solution to the system is:

• Response: (−4, −2)

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