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Which constants can be multiplied by the equations so one variable will be eliminated when the systems are added together?

5x + 13y = 232

12x + 7y = 218

The first equation can be multiplied by –13 and the second equation by 7 to eliminate y.
The first equation can be multiplied by 7 and the second equation by 13 to eliminate y.
The first equation can be multiplied by –12 and the second equation by 5 to eliminate x.
The first equation can be multiplied by 5 and the second equation by 12 to eliminate x.

11 answers

To eliminate one of the variables when adding the equations, we need to make the coefficients of that variable be opposites.

Let's analyze the two equations:

  1. \( 5x + 13y = 232 \)
  2. \( 12x + 7y = 218 \)

To eliminate \( y \):

  • The coefficient of \( y \) in the first equation is \( 13 \).
  • The coefficient of \( y \) in the second equation is \( 7 \).

To eliminate \( y \), we can multiply the first equation by \(-7\) (the coefficient of \( y \) in the second equation) and the second equation by \(13\) (the coefficient of \( y \) in the first equation):

  1. Multiply equation 1 by \(-7\): \[ -7(5x + 13y) = -7(232) \implies -35x - 91y = -1624 \]

  2. Multiply equation 2 by \(13\): \[ 13(12x + 7y) = 13(218) \implies 156x + 91y = 2834 \]

Now, if we add these modified equations, the \( y \) terms will cancel out.

Conclusion for \( y \):

  • First equation times \(-7\) and second equation times \(13\) can eliminate \( y \).

To eliminate \( x \):

  • The coefficient of \( x \) in the first equation is \( 5 \).
  • The coefficient of \( x \) in the second equation is \( 12 \).

To eliminate \( x \), we can multiply the first equation by \(-12\) and the second equation by \(5\) so that their coefficients become opposites:

  1. Multiply equation 1 by \(-12\): \[ -12(5x + 13y) = -12(232) \implies -60x - 156y = -2784 \]

  2. Multiply equation 2 by \(5\): \[ 5(12x + 7y) = 5(218) \implies 60x + 35y = 1090 \]

Now, adding these modified equations will eliminate \( x \).

Conclusion for \( x \):

  • First equation times \(-12\) and second equation times \(5\) can eliminate \( x \).

Summary of Options:

  1. The first equation can be multiplied by –13 and the second equation by 7 to eliminate \( y \). (This is incorrect based on calculations above.)
  2. The first equation can be multiplied by 7 and the second equation by 13 to eliminate \( y \). (This is correct.)
  3. The first equation can be multiplied by –12 and the second equation by 5 to eliminate \( x \). (This is correct.)
  4. The first equation can be multiplied by 5 and the second equation by 12 to eliminate \( x \). (This is incorrect based on calculations above.)

In conclusion, both the second and third options can be used as valid multipliers to eliminate the respective variables.

The first equation can be multiplied by –12 and the second equation by 5 to eliminate x.

Henrique began to solve a system of linear equations using the linear combination method. His work is shown below:

3(4x – 7y = 28) → 12x – 21y = 84
–2(6x – 5y = 31) → –12x + 10y = –62
12x – 21y = 84
+ –12x + 10y = –62
–11y = 22
y = –2
Complete the steps used to solve a system of linear equations by substituting the value of y into one of the original equations to find the value of x.

What is the solution to the system?

( , )

To find the solution to the system of equations using the value of \( y \) that Henrique computed, we will substitute \( y = -2 \) back into one of the original equations to find \( x \).

Let's start with the first equation from the original system: \[ 4x - 7y = 28 \]

Now, substitute \( y = -2 \): \[ 4x - 7(-2) = 28 \] \[ 4x + 14 = 28 \]

Next, isolate \( x \) by subtracting \( 14 \) from both sides: \[ 4x = 28 - 14 \] \[ 4x = 14 \]

Now, divide by \( 4 \): \[ x = \frac{14}{4} = \frac{7}{2} = 3.5 \]

Now we have the values of both \( x \) and \( y \):

  • \( x = 3.5 \)
  • \( y = -2 \)

Thus, the solution to the system of equations is: \[ \left( 3.5, -2 \right) \]

So the final answer is \( \boxed{(3.5, -2)} \).

times each equation by a number that produces opposite coefficients for x or y. 4x+5y=7
3x-2y=-12

To eliminate a variable in the system of equations when using the method of linear combinations, we multiply each equation by coefficients that will produce opposite coefficients for either \( x \) or \( y \).

Given Equations:

  1. \( 4x + 5y = 7 \) (Equation 1)
  2. \( 3x - 2y = -12 \) (Equation 2)

To eliminate \( x \):

  • The coefficient of \( x \) in Equation 1 is \( 4 \).
  • The coefficient of \( x \) in Equation 2 is \( 3 \).

To make the coefficients opposite, we can multiply:

  • Equation 1 by \( 3 \) (to match the coefficient of Equation 2 multiplied by 3).
  • Equation 2 by \( 4 \) (to match the coefficient of Equation 1 multiplied by 4).

Steps:

  1. Multiply Equation 1 by \( 3 \): \[ 3(4x + 5y) = 3(7) \implies 12x + 15y = 21 \]

  2. Multiply Equation 2 by \( 4 \): \[ 4(3x - 2y) = 4(-12) \implies 12x - 8y = -48 \]

Now, the modified equations are:

  1. \( 12x + 15y = 21 \)
  2. \( 12x - 8y = -48 \)

To eliminate \( x \):

Now we can eliminate \( x \) by subtracting the second equation from the first: \[ (12x + 15y) - (12x - 8y) = 21 - (-48) \]

This simplifies to: \[ 15y + 8y = 21 + 48 \] \[ 23y = 69 \] \[ y = 3 \]

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

Now we can substitute \( y = 3 \) back into one of the original equations to find \( x \). We'll use Equation 1: \[ 4x + 5(3) = 7 \] \[ 4x + 15 = 7 \] Subtract \( 15 \) from both sides: \[ 4x = 7 - 15 \] \[ 4x = -8 \] \[ x = -2 \]

Final Solution:

The solution to the system of equations is: \[ (x, y) = (-2, 3) \]

Thus, the solution is \( \boxed{(-2, 3)} \).

The linear combination method is applied to a system of equations:

(4x + 10y = 12)One-half
(10x + 25y = 30) – One-fifth
2x + 5y = 6
–2x – 5y = –6
0 = 0
What does 0 = 0 indicate about the solutions of the system?

There are 0 solutions to the system.
The solution to the system is (0, 0).
There are solutions to the system at the x- and y-intercepts.
There are infinitely many solutions to the system.

The result \(0 = 0\) indicates that the system of equations is dependent, meaning that both equations describe the same line. Therefore, every point on that line is a solution to the system.

In this case, we can conclude that:

  • There are infinitely many solutions to the system.

Thus, the correct answer is:

There are infinitely many solutions to the system.

Dawn has been using two bank accounts to save money for a car. The difference between account 1 and account 2 is $100. If she uses 3/8 of account 1 and 7/8 of account 2, Dawn will have a down payment of $2,000. Solve the system of equations to find the total amount of money Dawn has in each account.

A – B = 100

Three-eighthsA + Seven-eighthsB = 2,000

Dawn has $___in account 1 and $____in account 2.

Let's define the variables:

  • Let \( A \) be the amount of money in account 1.
  • Let \( B \) be the amount of money in account 2.

The system of equations is given as follows:

  1. \( A - B = 100 \) (Equation 1)
  2. \( \frac{3}{8}A + \frac{7}{8}B = 2000 \) (Equation 2)

Step 1: Solve the First Equation for \( A \)

From Equation 1, we can express \( A \) in terms of \( B \): \[ A = B + 100 \]

Step 2: Substitute into the Second Equation

Now we substitute \( A \) in Equation 2: \[ \frac{3}{8}(B + 100) + \frac{7}{8}B = 2000 \]

Step 3: Distribute and Combine Like Terms

Distributing: \[ \frac{3}{8}B + \frac{3}{8} \cdot 100 + \frac{7}{8}B = 2000 \] \[ \frac{3}{8}B + \frac{300}{8} + \frac{7}{8}B = 2000 \]

Now combine the \( B \) terms: \[ \left(\frac{3}{8}B + \frac{7}{8}B\right) + \frac{300}{8} = 2000 \] \[ \frac{10}{8}B + \frac{300}{8} = 2000 \]

Step 4: Eliminate the Fraction

To eliminate the fraction, multiply the entire equation by \( 8 \): \[ 10B + 300 = 16000 \]

Step 5: Solve for \( B \)

Now, isolate \( B \): \[ 10B = 16000 - 300 \] \[ 10B = 15700 \] \[ B = \frac{15700}{10} = 1570 \]

Step 6: Find \( A \)

Now that we have \( B \), we can find \( A \) using the expression \( A = B + 100 \): \[ A = 1570 + 100 = 1670 \]

Step 7: Conclusion

Dawn has:

  • $1670 in account 1
  • $1570 in account 2

Thus, the final answer is:

  • Dawn has $1670 in account 1 and $1570 in account 2.
A deli sells sliced meat and cheese. One customer purchases 4 pounds of meat and 5 pounds of cheese for a total of $30.50. A sandwich shop owner comes in and purchases 11 pounds of meat and 14 pounds of cheese for $84.50. The system of equations below represents the situation.

4x + 5y = 30.50

11x + 14y = 84.50

The variable x represents the
.

The variable y represents the
.

The deli charges $
.

To clarify the variables in your equations:

  1. The variable \( x \) typically represents the price per pound of meat.
  2. The variable \( y \) typically represents the price per pound of cheese.

Given the system of equations:

  • The first equation represents a customer's purchase: \[ 4x + 5y = 30.50 \] Here, \( 4x \) accounts for the cost of 4 pounds of meat, and \( 5y \) accounts for the cost of 5 pounds of cheese.

  • The second equation represents the sandwich shop owner's purchase: \[ 11x + 14y = 84.50 \] Here, \( 11x \) accounts for the cost of 11 pounds of meat, and \( 14y \) accounts for the cost of 14 pounds of cheese.

Solving the System of Equations:

Let's solve this system to find the values of \( x \) and \( y \).

Step 1: Use one equation to express \( y \) in terms of \( x \).

From the first equation: \[ 4x + 5y = 30.50 \] We can isolate \( y \): \[ 5y = 30.50 - 4x \] \[ y = \frac{30.50 - 4x}{5} \]

Step 2: Substitute \( y \) into the second equation.

Now substitute \( y \) into the second equation: \[ 11x + 14\left(\frac{30.50 - 4x}{5}\right) = 84.50 \]

Multiply everything by 5 to eliminate the fraction: \[ 55x + 14(30.50 - 4x) = 422.50 \] \[ 55x + 427 - 56x = 422.50 \] Combine like terms: \[ -1x + 427 = 422.50 \]

Step 3: Solve for \( x \).

Subtract 427 from both sides: \[ -1x = 422.50 - 427 \] \[ -1x = -4.50 \] \[ x = 4.50 \]

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

Substituting \( x = 4.50 \) back into the equation for \( y \): \[ y = \frac{30.50 - 4(4.50)}{5} \] \[ y = \frac{30.50 - 18}{5} \] \[ y = \frac{12.50}{5} = 2.50 \]

Conclusion:

  • The variable \( x \) represents the price per pound of meat, which is $4.50.
  • The variable \( y \) represents the price per pound of cheese, which is $2.50.
  • The deli charges $4.50 per pound for meat and $2.50 per pound for cheese.

Final response:

  • The deli charges $4.50 for meat and $2.50 for cheese.