Consider this system:
$\begin{array}{l}
3 x+\frac{1}{2} y=3 \\
6 x-y=2
\end{array}$
Which of the following operations would eliminate the $x$-terms if the two equations were added together afterward?
A. Multiply the first equation by -6.
B. Multiply the first equation by -2.
C. Multiply the first equation by 2.
D. Multiply the first equation by 6.
Answer (1)
Multiply the first equation by a variable m : m ( 3 x + 2 1 y ) = 3 m .
Add the modified equation to the second equation: ( 3 m + 6 ) x + ( 2 1 m − 1 ) y = 3 m + 2 .
Set the coefficient of x to zero: 3 m + 6 = 0 .
Solve for m : m = − 2 . The operation that eliminates the x -terms is to multiply the first equation by − 2 .
Explanation
Understanding the Problem We are given the following system of equations:
3 x + 2 1 y = 3 6 x − y = 2
Our goal is to find a number to multiply the first equation by such that when we add the modified first equation to the second equation, the x -terms are eliminated.
Setting up the Equation Let's denote the multiplier by m . We multiply the first equation by m :
m ( 3 x + 2 1 y ) = 3 m
This simplifies to:
3 m x + 2 1 m y = 3 m
Now, we add the modified first equation to the second equation:
( 3 m x + 2 1 m y ) + ( 6 x − y ) = 3 m + 2
Combining like terms, we get:
( 3 m + 6 ) x + ( 2 1 m − 1 ) y = 3 m + 2
To eliminate the x -terms, the coefficient of x must be zero. Therefore, we need to solve the equation:
3 m + 6 = 0
Solving for the Multiplier Solving for m :
3 m = − 6
m = 3 − 6
m = − 2
So, multiplying the first equation by -2 will eliminate the x -terms when the two equations are added together.
Final Answer Therefore, the operation that eliminates the x -terms is to multiply the first equation by -2.
Examples
Consider a scenario where you are trying to balance a chemical equation. You might need to multiply one equation by a certain factor so that when you add it to another equation, a particular element is eliminated, helping you solve for the remaining variables. This is analogous to eliminating x-terms in a system of equations to solve for y, and then substituting back to find x. This technique is widely used in various fields, including chemistry, physics, and engineering, to simplify complex problems and find solutions more efficiently.
