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In order to solve the following system of equations by addition, which of the following could you do before adding the equations so that one variable will be eliminated when you add them?

[tex]\[ \begin{array}{l}
4x - 2y = 7 \\
3x - 3y = 15
\end{array} \][/tex]

A. Multiply the top equation by 3 and the bottom equation by 2

B. Multiply the top equation by -3 and the bottom equation by 2

C. Multiply the top equation by [tex]\(\frac{1}{3}\)[/tex]

D. Multiply the top equation by 3 and the bottom equation by 4


Sagot :

To solve the given system of equations by addition and to eliminate one of the variables, we need the coefficients of that variable in both equations to be equal in magnitude but opposite in sign.

The given system of equations is:

[tex]\[ \begin{array}{l} 4x - 2y = 7 \quad \text{(Equation 1)} \\ 3x - 3y = 15 \quad \text{(Equation 2)} \end{array} \][/tex]

One effective method to eliminate a variable is to manipulate the equations such that the coefficients of [tex]\( y \)[/tex] (or [tex]\( x \)[/tex]) in both equations are the same in magnitude but opposite in sign.

Step-by-step:
1. Notice that the coefficients of [tex]\( y \)[/tex] in Equation 1 and Equation 2 are [tex]\(-2\)[/tex] and [tex]\(-3\)[/tex], respectively.
2. To make the coefficients of [tex]\( y \)[/tex] equal in magnitude but opposite in sign, we can find common multiples.

Let’s consider multiplication to achieve this:
- If we multiply Equation 1 by 3, the coefficient of [tex]\( y \)[/tex] in Equation 1 will become [tex]\( \-6\)[/tex] (because [tex]\(3 \times -2 = -6\)[/tex]).
- If we multiply Equation 2 by 2, the coefficient of [tex]\( y \)[/tex] in Equation 2 will also become [tex]\( -6\)[/tex] (because [tex]\(2 \times -3 = -6\)[/tex]).

Therefore, by multiplying:
- Equation 1 by 3: [tex]\( 3 \times (4x - 2y) = 3 \times 7 \)[/tex]
[tex]\[ 12x - 6y = 21 \quad \text{(New Equation 1)} \][/tex]

- Equation 2 by 2: [tex]\( 2 \times (3x - 3y) = 2 \times 15 \)[/tex]
[tex]\[ 6x - 6y = 30 \quad \text{(New Equation 2)} \][/tex]

When these new equations are added, the [tex]\( y \)[/tex]-terms will cancel each other:

[tex]\[ (12x - 6y) + (6x - 6y) = 21 + 30 \][/tex]
[tex]\[ 18x = 51 \][/tex]

Thus, the step necessary to set up the system for addition to eliminate [tex]\( y \)[/tex] involves:

Multiply the top equation by 3 (to make [tex]\(-2y\)[/tex] into [tex]\(-6y\)[/tex]) and multiply the bottom equation by 2 (to make [tex]\(-3y\)[/tex] into [tex]\(-6y\)[/tex]).

So option A. Multiply the top equation by 3 and the bottom equation by 2 is the correct step required before adding the equations to eliminate one variable, [tex]\( y \)[/tex].