Finding the Correct Value of k

Oct 13, 2025 by pinkandtrash.com 54 views

Introduction

In algebra, a system of equations is a set of two or more equations that are solved simultaneously to find the values of the variables. When dealing with a system of equations, it is often necessary to eliminate one of the variables in order to solve for the other. In this article, we will examine a system of equations and determine which number can be multiplied by each equation so that when the two equations are added together, the y-term is eliminated.

The System of Equations

The system of equations we will be examining is:

\frac{3}{4} x + \frac{2}{3} y = 6

\frac{5}{8} x + \frac{3}{4} y = 7

Eliminating the y-term

To eliminate the y-term, we need to multiply each equation by a number such that when the two equations are added together, the y-term is eliminated. Let's call the number we need to multiply the first equation by "k". Then, the first equation becomes:

k \left( \frac{3}{4} x + \frac{2}{3} y \right) = k(6)

\frac{3k}{4} x + \frac{2k}{3} y = 6k

Similarly, the second equation becomes:

k \left( \frac{5}{8} x + \frac{3}{4} y \right) = k(7)

\frac{5k}{8} x + \frac{3k}{4} y = 7k

Finding the Value of k

To eliminate the y-term, we need to find a value of k such that when the two equations are added together, the y-term is eliminated. This means that the coefficients of the y-term in both equations must be equal and opposite. In other words, we need to find a value of k such that:

\frac{2k}{3} = -\frac{3k}{4}

Solving for k, we get:

\frac{2k}{3} = -\frac{3k}{4}

8k = -9k

17k = 0

k = 0

However, k cannot be 0, as this would mean that the y-term is not eliminated. Therefore, we need to find another value of k.

Finding the Correct Value of k

Let's try to find a value of k such that the coefficients of the y-term in both equations are equal and opposite. We can do this by multiplying the first equation by a number that will make the coefficient of the y-term in the first equation equal to the negative of the coefficient of the y-term in the second equation.

The coefficient of the y-term in the first equation is $\frac{2}{3}$ . The coefficient of the y-term in the second equation is $\frac{3}{4}$ . To make the coefficient of the y-term in the first equation equal to the negative of the coefficient of the y-term in the second equation, we need to multiply the first equation by a number that will make $\frac{2}{3}$ equal to $-\frac{3}{4}$ .

However, we can't find a number that will make $\frac{2}{3}$ equal to $-\frac{3}{4}$ . But we can find a number that will make $\frac{2}{3}$ equal to $\frac{3}{4}$ .

To make $\frac{2}{3}$ equal to $\frac{3}{4}$ , we need to multiply the first equation by a number that will make the denominator of $\frac{2}{3}$ equal to the denominator of $\frac{3}{4}$ . The denominator of $\frac{2}{3}$ is 3, and the denominator of $\frac{3}{4}$ is 4. Therefore, we need to multiply the first equation by a number that will make 3 equal to 4.

However, we can't find a number that will make 3 equal to 4. But we can find a number that will make 3 equal to 12.

To make 3 equal to 12, we need to multiply the first equation by 4.

Therefore, the correct value of k is 4.

The Corrected System of Equations

The corrected system of equations is:

4 \left( \frac{3}{4} x + \frac{2}{3} y \right) = 4(6)

\frac{12}{4} x + \frac{8}{3} y = 24

\frac{12}{4} x + \frac{8}{3} y = 24

\frac{5}{8} x + \frac{3}{4} y = 7

Adding the Two Equations

To add the two equations, we need to make the coefficients of the y-term in both equations equal. We can do this by multiplying the second equation by a number that will make the coefficient of the y-term in the second equation equal to the negative of the coefficient of the y-term in the first equation.

The coefficient of the y-term in the first equation is $\frac{8}{3}$ . The coefficient of the y-term in the second equation is $\frac{3}{4}$ . To make the coefficient of the y-term in the second equation equal to the negative of the coefficient of the y-term in the first equation, we need to multiply the second equation by a number that will make $\frac{3}{4}$ equal to $-\frac{8}{3}$ .

However, we can't find a number that will make $\frac{3}{4}$ equal to $-\frac{8}{3}$ . But we can find a number that will make $\frac{3}{4}$ equal to $\frac{8}{3}$ .

To make $\frac{3}{4}$ equal to $\frac{8}{3}$ , we need to multiply the second equation by a number that will make the denominator of $\frac{3}{4}$ equal to the denominator of $\frac{8}{3}$ . The denominator of $\frac{3}{4}$ is 4, and the denominator of $\frac{8}{3}$ is 3. Therefore, we need to multiply the second equation by a number that will make 4 equal to 3.

However, we can't find a number that will make 4 equal to 3. But we can find a number that will make 4 equal to 12.

To make 4 equal to 12, we need to multiply the second equation by 3.

Therefore, the corrected system of equations is:

\frac{12}{4} x + \frac{8}{3} y = 24

3 \left( \frac{5}{8} x + \frac{3}{4} y \right) = 3(7)

\frac{15}{8} x + \frac{9}{4} y = 21

Adding the Two Equations

To add the two equations, we need to make the coefficients of the x-term in both equations equal. We can do this by multiplying the first equation by a number that will make the coefficient of the x-term in the first equation equal to the coefficient of the x-term in the second equation.

The coefficient of the x-term in the first equation is $\frac{12}{4}$ . The coefficient of the x-term in the second equation is $\frac{15}{8}$ . To make the coefficient of the x-term in the first equation equal to the coefficient of the x-term in the second equation, we need to multiply the first equation by a number that will make $\frac{12}{4}$ equal to $\frac{15}{8}$ .

However, we can't find a number that will make $\frac{12}{4}$ equal to $\frac{15}{8}$ . But we can find a number that will make $\frac{12}{4}$ equal to $\frac{15}{8}$ .

To make $\frac{12}{4}$ equal to $\frac{15}{8}$ , we need to multiply the first equation by a number that will make the denominator of $\frac{12}{4}$ equal to the denominator of $\frac{15}{8}$ . The denominator of $\frac{12}{4}$ is 4, and the denominator of $\frac{15}{8}$ is 8. Therefore, we need to multiply the first equation by a number that will make 4 equal to 8.

However, we can't find a number that will make 4 equal to 8. But we can find a number that will make 4 equal to 16.

To make 4 equal to 16, we need to multiply the first equation by 4.

Therefore, the corrected system of equations is:

4 \left( \frac{12}{4} x + \frac{8}{3} y \right) = 4(24)

\frac{48}{4} x + \frac{32}{3} y = 96

\frac{48}{4} x + \frac{32}{3} y = 96

\frac{15}{8} x + \frac{9}{4} y = 21

Adding the Two Equations

Q: What is the main goal of eliminating the y-term in a system of equations?

A: The main goal of eliminating the y-term in a system of equations is to simplify the system and make it easier to solve for the values of the variables.

Q: How do you eliminate the y-term in a system of equations?

A: To eliminate the y-term in a system of equations, you need to multiply each equation by a number such that when the two equations are added together, the y-term is eliminated. This means that the coefficients of the y-term in both equations must be equal and opposite.

Q: What is the correct value of k to eliminate the y-term in the given system of equations?

A: The correct value of k to eliminate the y-term in the given system of equations is 4.

Q: How do you add the two equations together to eliminate the y-term?

A: To add the two equations together to eliminate the y-term, you need to make the coefficients of the x-term in both equations equal. You can do this by multiplying the first equation by a number that will make the coefficient of the x-term in the first equation equal to the coefficient of the x-term in the second equation.

Q: What is the final result of adding the two equations together to eliminate the y-term?

A: The final result of adding the two equations together to eliminate the y-term is a new equation with only one variable, x.

Q: How do you solve for the value of x in the new equation?

A: To solve for the value of x in the new equation, you can use algebraic methods such as substitution or elimination.

Q: What is the importance of eliminating the y-term in a system of equations?

A: Eliminating the y-term in a system of equations is important because it simplifies the system and makes it easier to solve for the values of the variables. It also helps to reduce the number of variables in the system, making it easier to understand and analyze.

Q: Can you provide an example of a system of equations where eliminating the y-term is necessary?

A: Yes, here is an example of a system of equations where eliminating the y-term is necessary:

\frac{3}{4} x + \frac{2}{3} y = 6

\frac{5}{8} x + \frac{3}{4} y = 7

In this system of equations, eliminating the y-term is necessary to solve for the values of x and y.

Q: How do you know when to eliminate the y-term in a system of equations?

A: You know when to eliminate the y-term in a system of equations when the coefficients of the y-term in both equations are not equal and opposite. In this case, you need to multiply each equation by a number such that when the two equations are added together, the y-term is eliminated.

Q: Can you provide a step-by-step guide on how to eliminate the y-term in a system of equations?

A: Yes, here is a step-by-step guide on how to eliminate the y-term in a system of equations:

Multiply each equation by a number such that when the two equations are added together, the y-term is eliminated.
Add the two equations together to eliminate the y-term.
Simplify the resulting equation to get a new equation with only one variable.
Solve for the value of the variable in the new equation using algebraic methods such as substitution or elimination.

By following these steps, you can eliminate the y-term in a system of equations and simplify the system to make it easier to solve for the values of the variables.