Question

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.

Answer 1

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?

( , )

Answer 2

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

Answer 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.

Answer 4

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.

Answer 5

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 $
.

Answer 6

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.

Answer 7

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 \( \boxed{(3.5, -2)} \).

Answer 8

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 \( \boxed{(-2, 3)} \).

Answer 9

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.**

Answer 10

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.

Answer 11

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.