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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

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
4 x plus 14 y equals 28 and 8 x plus 10 y equals 10

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

9 answers

To determine which of the given options shows a system of equations equivalent to the original system \(2x + 7y = 14\) and \(4x + 5y = 10\), we can manipulate both equations and check for equivalence.

  1. The first equation \(2x + 7y = 14\) can be multiplied by any non-zero constant to yield an equivalent equation. For example, multiplying by 2 gives: \[ 4x + 14y = 28 \quad \text{(equivalent to the original equation)} \]

  2. The second equation \(4x + 5y = 10\) can also be multiplied or manipulated. For example, dividing by 2 gives: \[ 2x + 2.5y = 5 \quad \text{(not shown in the options)} \]

Now, let's check the options:

  1. First option: \(2x + 7y = 14\) and (2x + 5y = 5}

    • \(2x + 7y = 14\) is equivalent, but \(2x + 5y = 5\) is different from \(4x + 5y = 10\). Not equivalent.

  2. Second option: \(2x + 7y = 14\) and \(8x + 10y = 20\)

    • \(8x + 10y = 20\) can be derived by multiplying \(4x + 5y = 10\) by 2. This option is equivalent to the original equations.

  3. Third option: \(4x + 14y = 28\) and \(8x + 10y = 10\)

    • While \(4x + 14y = 28\) is equivalent, \(8x + 10y = 10\) is not equivalent to \(4x + 5y = 10\) when manipulated. Not equivalent.

  4. Fourth option: \( 4x + 14y = 14\) and \( 4x + 5y = 10\)

    • Here, \(4x + 14y = 14\) is not equivalent to \(4x + 7y = 14\). Not equivalent.

The only option that provides an equivalent system of equations is:

2. \(2x + 7y = 14\) and \(8x + 10y = 20\)

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

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).
The intercepts of both 4 x plus 5 y equals 10 and 8 x plus 10 y equals 20 are left parenthesis 0 comma 2.5 right parenthesis and left parenthesis 2 comma 0 right parenthesis .

All equations share the point (0,2).

To determine why the system of equations \(2x + 7y = 14\) and \(4x + 5y = 10\) is equivalent to the system \(4x + 14y = 28\) and \(8x + 10y = 20\), we need to look at the transformations applied to create equivalent equations.

  1. Multiplying the first equation \(2x + 7y = 14\) by 2 gives: \[ 2(2x + 7y) = 2(14) \implies 4x + 14y = 28 \]

  2. Multiplying the second equation \(4x + 5y = 10\) by 2 gives: \[ 2(4x + 5y) = 2(10) \implies 8x + 10y = 20 \]

Thus, both equations in the original system can be multiplied by the same constant (2 in this case) to yield the new system, which makes them equivalent.

Now let's evaluate the provided responses in this context:

  1. First option: "Both equations, 2x + 7y = 14 and 4x + 5y = 10, can be multiplied by 12 to produce the equivalent system of equations."

    • This is incorrect because multiplying by 12 will not yield the new system.

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

    • This is incorrect because multiplying by \(\frac{1}{2}\) would produce a different system, not the equivalent one presented.

  3. Third option: "The intercepts of the equations 2x + 7y = 14 and 4x + 14y = 28 are (0,2) and (7,0)."

    • This is incorrect; these equations do not share the same intercepts.

  4. Fourth option: "The intercepts of the equations 2x + 7y = 14 and 4x + 14y = 28 are (0,2) and (7,0)."

    • This is also incorrect, as noted above.

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

    • This is not relevant to why the original equations are equivalent.

  6. Sixth option: "All equations share the point (0,2)."

    • This could potentially be true, but it is not the best explanation for the equivalence of the systems based on multiplication.

The proper reasoning lies in recognizing that the transformations applied to the original equations through multiplication by the same non-zero constant (here by 2) yield equivalent equations. Unfortunately, that reasoning does not directly appear in the provided options.

However, if one option properly states that the first equations can be multiplied by 2 to produce the equivalent system, that would be the accurate response. Since none of the options reflect this, the precise logical explanation is missing, but the best understanding of equivalence revolves around the scaling of the equations.

you have to choose one

Given the need to select one of the provided options, the best choice based on the rationale should be:

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

Although this option is not entirely correct because multiplying by \(\frac{1}{2}\) does not give the new system (it would yield a different system), it focuses on a transformation concept which is relevant in discussing the equivalence of equations. However, note that the correct transformation involves multiplying by 2, not \(\frac{1}{2}\).

If you must choose strictly from the provided phrases, this would still be a reference to the concept of transformation between the equations, even if it's not accurate as stated.

choose only one

Given your prompt and the options provided, the best choice is:

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

This option at least acknowledges the concept of manipulating the equations, even though multiplying by \(\frac{1}{2}\) specifically does not produce the equivalent system. It's important to note that no option perfectly captures the correct transformation, but this is the closest relevant choice.

Solve the following system of equations using the elimination method.

3x−4y=−4, x+3y=−10

(1 point)
Responses

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

(−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

To solve the system of equations using the elimination method, we'll work with the following equations:

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

Step 1: Eliminate one variable

To eliminate \(x\) from the equations, we can first multiply Equation 2 by 3 (so that the coefficients of \(x\) in both equations will be the same):

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

This simplifies to: \[ 3x + 9y = -30 \quad (Equation 3) \]

Now our system is:

  1. \(3x - 4y = -4\)
  2. \(3x + 9y = -30\)

Step 2: Subtract the first equation from the second

Now we can subtract Equation 1 from Equation 3:

\[ (3x + 9y) - (3x - 4y) = -30 - (-4) \]

Simplifying this gives: \[ 3x + 9y - 3x + 4y = -30 + 4 \] \[ 13y = -26 \]

Step 3: Solve for \(y\)

Now, divide both sides by 13:

\[ y = -2 \]

Step 4: Substitute \(y\) back into one of the original equations

We can now substitute \(y = -2\) back into Equation 2 to solve for \(x\):

\[ x + 3(-2) = -10 \] \[ x - 6 = -10 \] \[ x = -10 + 6 \] \[ x = -4 \]

Final Result

The solution to the system of equations is:

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

Thus, the answer is:

(−4,−2)