Multiply Each Equation By A Number That Produces Opposite Coefficients For X X X Or Y Y Y .1. 2 5 X + 6 Y = − 10 \frac{2}{5} X + 6y = -10 5 2 ​ X + 6 Y = − 10 2. − 2 X − 2 Y = 40 -2x - 2y = 40 − 2 X − 2 Y = 40 Determine What Number To Multiply The First Equation By To Form Opposite Terms

Introduction

In algebra, we often encounter systems of linear equations that require us to manipulate the equations to isolate the variables. One common technique is to multiply each equation by a number that produces opposite coefficients for the variables. This technique is essential in solving systems of linear equations and is a fundamental concept in mathematics. In this article, we will explore how to multiply equations to form opposite terms and apply this technique to two given equations.

Understanding Opposite Coefficients

Opposite coefficients refer to the coefficients of the variables in an equation that have opposite signs. For example, in the equation $2x - 3y = 5$, the coefficients of $x$ and $y$ are $2$ and $-3$, respectively. These coefficients are opposite because they have opposite signs.

Multiplying Equations to Form Opposite Terms

To multiply an equation by a number that produces opposite coefficients for the variables, we need to find a number that, when multiplied by the coefficient of one variable, results in the negative of the coefficient of the other variable. Let's consider the first equation:

$\frac{2}{5} x + 6y = -10$

We want to find a number that, when multiplied by $\frac{2}{5}$, results in the negative of $6$. This number is $-\frac{5}{2}$, because:

$-\frac{5}{2} \times \frac{2}{5} = -1$

However, we need to find a number that, when multiplied by $6$, results in the negative of $\frac{2}{5}$. This number is $-\frac{1}{30}$, because:

$-\frac{1}{30} \times 6 = -\frac{1}{5}$

But we are looking for a number that will give us the opposite of $\frac{2}{5}$ when multiplied by $6$. This number is actually $-\frac{1}{30}$ multiplied by $-5$, which is $\frac{1}{6}$.

However, we can multiply the entire equation by $-5$ to get:

$-2x - 30y = 50$

Now, we can see that the coefficients of $x$ and $y$ are $-2$ and $-30$, respectively. These coefficients are opposite because they have opposite signs.

Applying the Technique to the Second Equation

Now, let's consider the second equation:

$-2x - 2y = 40$

We want to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is $-1$, because:

$-1 \times -2 = 2$

However, we need to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is actually $-1$.

Conclusion

In conclusion, multiplying equations to form opposite terms is a fundamental technique in algebra that requires us to find a number that, when multiplied by the coefficient of one variable, results in the negative of the coefficient of the other variable. By applying this technique to the two given equations, we can see that the coefficients of the variables can be made to have opposite signs by multiplying the equations by a suitable number. This technique is essential in solving systems of linear equations and is a fundamental concept in mathematics.

Example 1: Multiplying the First Equation

Let's consider the first equation:

$\frac{2}{5} x + 6y = -10$

We want to find a number that, when multiplied by $\frac{2}{5}$, results in the negative of $6$. This number is $-\frac{5}{2}$, because:

$-\frac{5}{2} \times \frac{2}{5} = -1$

However, we need to find a number that, when multiplied by $6$, results in the negative of $\frac{2}{5}$. This number is actually $-\frac{1}{30}$ multiplied by $-5$, which is $\frac{1}{6}$.

However, we can multiply the entire equation by $-5$ to get:

$-2x - 30y = 50$

Now, we can see that the coefficients of $x$ and $y$ are $-2$ and $-30$, respectively. These coefficients are opposite because they have opposite signs.

Example 2: Multiplying the Second Equation

Let's consider the second equation:

$-2x - 2y = 40$

We want to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is $-1$, because:

$-1 \times -2 = 2$

However, we need to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is actually $-1$.

Step-by-Step Solution

To solve the system of linear equations, we can follow these steps:

1. Multiply the first equation by $-5$ to get:

$-2x - 30y = 50$

2. Multiply the second equation by $-1$ to get:

$2x + 2y = -40$

3. Add the two equations to eliminate the variable $x$:

$-28y = 10$

4. Solve for $y$:

$y = -\frac{10}{28}$

5. Substitute the value of $y$ into one of the original equations to solve for $x$:

$\frac{2}{5} x + 6(-\frac{10}{28}) = -10$

6. Solve for $x$:

$x = -\frac{25}{2}$

Final Answer

The final answer is:

$x = -\frac{25}{2}$

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Introduction

In our previous article, we explored how to multiply equations to form opposite terms. This technique is essential in solving systems of linear equations and is a fundamental concept in mathematics. In this article, we will answer some common questions related to multiplying equations to form opposite terms.

Q: What is the purpose of multiplying equations to form opposite terms?

A: The purpose of multiplying equations to form opposite terms is to make the coefficients of the variables have opposite signs. This allows us to eliminate one of the variables by adding or subtracting the two equations.

Q: How do I know which number to multiply the equation by?

A: To determine which number to multiply the equation by, you need to find a number that, when multiplied by the coefficient of one variable, results in the negative of the coefficient of the other variable.

Q: What if I multiply the equation by a number that doesn't produce opposite coefficients?

A: If you multiply the equation by a number that doesn't produce opposite coefficients, you will not be able to eliminate one of the variables by adding or subtracting the two equations. In this case, you will need to try a different number.

Q: Can I multiply the equation by a fraction?

A: Yes, you can multiply the equation by a fraction. However, you need to make sure that the fraction is simplified and that the numerator and denominator are not both zero.

Q: What if I multiply the equation by a negative number?

A: If you multiply the equation by a negative number, the signs of the coefficients will change. For example, if you multiply the equation by $-2$, the coefficients will become $-2$ and $2$.

Q: Can I multiply the equation by a decimal number?

A: Yes, you can multiply the equation by a decimal number. However, you need to make sure that the decimal number is simplified and that the numerator and denominator are not both zero.

Q: What if I multiply the equation by a mixed number?

A: If you multiply the equation by a mixed number, you need to convert the mixed number to an improper fraction before multiplying.

Q: Can I multiply the equation by a variable?

A: No, you cannot multiply the equation by a variable. The equation must be multiplied by a number, not a variable.

Q: What if I multiply the equation by a zero?

A: If you multiply the equation by a zero, the equation will become zero, and you will not be able to solve for the variables.

Q: Can I multiply the equation by a negative fraction?

A: Yes, you can multiply the equation by a negative fraction. However, you need to make sure that the fraction is simplified and that the numerator and denominator are not both zero.

Q: What if I multiply the equation by a negative decimal number?

A: If you multiply the equation by a negative decimal number, the signs of the coefficients will change. For example, if you multiply the equation by $-0.5$, the coefficients will become $-0.5$ and $0.5$.

Q: Can I multiply the equation by a negative mixed number?

A: Yes, you can multiply the equation by a negative mixed number. However, you need to convert the mixed number to an improper fraction before multiplying.

Conclusion

In conclusion, multiplying equations to form opposite terms is a fundamental technique in algebra that requires us to find a number that, when multiplied by the coefficient of one variable, results in the negative of the coefficient of the other variable. By answering these common questions, we can better understand how to multiply equations to form opposite terms and apply this technique to solve systems of linear equations.

Example 1: Multiplying the First Equation

Let's consider the first equation:

$\frac{2}{5} x + 6y = -10$

We want to find a number that, when multiplied by $\frac{2}{5}$, results in the negative of $6$. This number is $-\frac{5}{2}$, because:

$-\frac{5}{2} \times \frac{2}{5} = -1$

However, we need to find a number that, when multiplied by $6$, results in the negative of $\frac{2}{5}$. This number is actually $-\frac{1}{30}$ multiplied by $-5$, which is $\frac{1}{6}$.

However, we can multiply the entire equation by $-5$ to get:

$-2x - 30y = 50$

Now, we can see that the coefficients of $x$ and $y$ are $-2$ and $-30$, respectively. These coefficients are opposite because they have opposite signs.

Example 2: Multiplying the Second Equation

Let's consider the second equation:

$-2x - 2y = 40$

We want to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is $-1$, because:

$-1 \times -2 = 2$

However, we need to find a number that, when multiplied by $-2$, results in the negative of $-2$. This number is actually $-1$.

Step-by-Step Solution

To solve the system of linear equations, we can follow these steps:

1. Multiply the first equation by $-5$ to get:

$-2x - 30y = 50$

2. Multiply the second equation by $-1$ to get:

$2x + 2y = -40$

3. Add the two equations to eliminate the variable $x$:

$-28y = 10$

4. Solve for $y$:

$y = -\frac{10}{28}$

5. Substitute the value of $y$ into one of the original equations to solve for $x$:

$\frac{2}{5} x + 6(-\frac{10}{28}) = -10$

6. Solve for $x$:

$x = -\frac{25}{2}$

Final Answer

The final answer is:

$x = -\frac{25}{2}$

$y = -\frac{10}{28}$