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

Responses

infinite solutions
infinite solutions

no solution
no solution

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

(2−1)

11 answers

To solve the system of equations using the substitution method, we first express one variable in terms of the other.

The given system is:

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

We can start by solving Equation 1 for \( y \):

\[ 3y = 4 - 2x \]

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

Now, we can substitute this expression for \( y \) into Equation 2. However, first, notice that Equation 2 can be simplified directly by dividing all terms by -2:

\[ 2x + 3y = 4 \]

This shows that both equations represent the same line because Equation 1 and the simplified Equation 2 are identical.

Since both equations are the same, there are infinite solutions along the line defined by \( 2x + 3y = 4 \).

Thus, the correct response to the system of equations is:

infinite solutions.

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

infinite solutions
infinite solutions

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

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

no solution

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

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

We can start by substituting the expression for \( x \) from Equation 1 into Equation 2.

Substituting \( x = y - 4 \) into Equation 2:

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

Combine like terms:

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

Now, add 4 to both sides:

\[ 9y = 6 \]

Then, divide by 9:

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

Now that we have \( y \), we can substitute it back into Equation 1 to find \( x \):

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

To express it with a common denominator:

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

So the solution to the system is:

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

Now, we compare this to the given responses. Note that it seems there are no exact matches with my calculation.

Thus, the answer should be:

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

Verify whether any of the given options reflect this solution accurately:

  • \( (23, -313) \)
  • \( (-313, 23) \)
  • \( \text{no solution} \)

None of these choices are correct based on the calculations. Please double-check your options or provide the correct choices if there's an error. Would you like to confirm or correct the response options?

yes

Great! Please provide the correct options or confirm any modifications to the existing options based on our solution of

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

If possible, I can help verify the correctness of any new options you provide!

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

infinite solutions
infinite solutions

(2/3,−3/13)
left parenthesis Start Fraction 2 over 3 End Fraction comma negative 3 Start Fraction 1 over 3 End Fraction right parenthesis

(−3/13,2/3)
left parenthesis negative 3 Start Fraction 1 over 3 End Fraction comma Start Fraction 2 over 3 End Fraction right parenthesis

no solution

Let's solve the system of equations again using the substitution method to ensure accuracy.

The equations are:

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

Step 1: Substitute \( x \) in Equation 2

From Equation 1, we know:

\[ x = y - 4 \]

Now substitute \( x \) in Equation 2:

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

Step 2: Combine like terms

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

Step 3: Solve for \( y \)

Add 4 to both sides:

\[ 9y = 6 \]

Now divide by 9:

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

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

Now use the value of \( y \) to find \( x \):

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

Now we have:

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

However, it seems there is a need to verify if we had made an error or if the answer relates to those options you provided.

Check the Possible Responses:

  1. infinite solutions
  2. \((\frac{2}{3}, -\frac{3}{13})\)
  3. \((- \frac{3}{13}, \frac{2}{3})\)
  4. no solution

Solution Misalignment:

The solutions we've found do not match any of the specified options:

  • Our calculated solution is \( \left( -\frac{10}{3}, \frac{2}{3} \right) \).

Verification of Given Responses:

  1. Infinite solutions would apply were the lines coincident, which isn't the case here.
  2. No solution would apply if the lines were parallel, which isn't the case here either.
  3. Upon checking: Neither of \((\frac{2}{3}, -\frac{3}{13})\) nor \((- \frac{3}{13}, \frac{2}{3})\) match our solution.

Conclusion:

The calculated solution \( \left( -\frac{10}{3}, \frac{2}{3} \right) \) does not match any of the provided options. There might be either a transcription error with the equations or the solution options.

Would you like to verify the problem statement again or see if the equations were appropriately transcribed?

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

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

infinite solutions
infinite solutions

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

no solution

Let's solve the system of equations using the substitution method.

The given system is:

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

Step 1: Solve Equation 1 for \( y \)

From Equation 1, we can express \( y \) in terms of \( x \):

\[ y = -2 - 2.5x \]

Step 2: Substitute \( y \) in Equation 2

Now substitute \( y \) in Equation 2:

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

Distributing the \( 2 \):

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

Step 3: Combine like terms

Combine the \( x \) terms:

\[ -2x - 4 = 0 \]

Step 4: Solve for \( x \)

Add \( 4 \) to both sides:

\[ -2x = 4 \]

Now divide by \(-2\):

\[ x = -2 \]

Step 5: Substitute \( x \) back to find \( y \)

Now substitute \( x = -2 \) back into the equation for \( y \):

\[ y = -2 - 2.5(-2) \] \[ y = -2 + 5 = 3 \]

Conclusion

The solution to the system of equations is:

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

Compare with the Given Responses:

  1. \( (3, -2) \)
  2. infinite solutions
  3. (-2, 3)
  4. no solution

The solution we found matches option (-2, 3).

Thus, the correct response is:

\[ \boxed{(-2, 3)} \]

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

To solve the linear-quadratic system given by the equations \(y = x^2 - 4x + 8\) and \(y = 3x - 4\) using the substitution method, we can follow these steps:

  1. Set the two equations equal to each other:

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

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

  1. Rearrange the equation:

Move all terms to one side:

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

This simplifies to:

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

  1. Factor the quadratic equation:

Now, we can factor the quadratic:

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

  1. Solve for \(x\):

Setting each factor to zero gives us the solutions:

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

  1. Find corresponding \(y\) values:

Now we substitute each \(x\) value back into either original equation to find \(y\).

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

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

Final Solutions:

The solutions to the system are:

  1. \( (3, 5) \)
  2. \( (4, 8) \)

Compare with Given Responses:

  1. \( (4, 8) \)
  2. \( (3, 5) \)
  3. \( (4, 8) \) and \( (3, 5) \)
  4. \( (-4, -16) \) and -3

Conclusion:

The correct response is:

\[ \boxed{(4, 8) \text{ and } (3, 5)} \]

This corresponds to option (4, 8) and (3, 5).

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