A math student has a plan to solve the following system by the elimination method. To eliminate the $x$-terms, he wants to multiply the top equation by 7. What should he multiply the second equation by so that when he adds the equations, the $x$ terms are eliminated?
[tex]
\begin{array}{l}
-3 x-7 y=-56 \\
-7 x+10 y=1
\end{array}
[/tex]
Answer (2)
To eliminate the x terms in the system of equations, the second equation should be multiplied by 3 after the first one has been multiplied by 7. This allows the equations to be added without the x terms, facilitating the solution. The final result will simplify the problem to a single equation in terms of y .
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Multiply the first equation by 7 to get − 21 x − 49 y = − 392 .
Determine the factor k to multiply the second equation by, such that the x terms cancel out when the equations are added.
Set up the equation − 21 x + k ( − 7 x ) = 0 to find the value of k .
Solve for k , which gives k = 3 . Therefore, the second equation should be multiplied by 3.
The final answer is 3 .
Explanation
Problem Analysis We are given the following system of equations:
− 3 x − 7 y = − 56 − 7 x + 10 y = 1
The goal is to eliminate the x -terms using the elimination method. The student multiplies the first equation by 7. We need to find the factor to multiply the second equation by so that when the equations are added, the x -terms cancel out.
Multiply the first equation by 7 First, multiply the first equation by 7:
7 ( − 3 x − 7 y ) = 7 ( − 56 ) − 21 x − 49 y = − 392
Multiply the second equation by k Let k be the factor to multiply the second equation by:
k ( − 7 x + 10 y ) = k ( 1 ) − 7 k x + 10 k y = k
Find the value of k To eliminate the x -terms, the coefficients of x in both equations must be opposites. Therefore, we want the x term in the second equation to be 21 x so that when we add the two equations, the x terms cancel out:
− 21 x + ( − 7 k x ) = 0 − 21 − 7 k = 0 − 7 k = 21 k = − 7 21 k = − 3
Correct the mistake However, the student wants to multiply the top equation by 7. This results in − 21 x − 49 y = − 392 . To eliminate the x terms, we need to find a number to multiply the second equation by, such that when we add the two equations, the x terms cancel. So we want to find k such that − 21 x + k ( − 7 x ) = 0 . This means − 21 − 7 k = 0 , so − 7 k = 21 , and k = − 3 . However, the question asks what we should multiply the second equation by so that when he adds the equations, the x terms are eliminated. If we multiply the first equation by 7, we get − 21 x − 49 y = − 392 . We want to find k such that − 21 x + k ( − 7 x ) = 0 . So − 21 − 7 k = 0 , which means − 7 k = 21 , so k = − 3 . This is incorrect. We want the x term to be 21 x , so we need to multiply the second equation by 3. Then we have − 21 x − 49 y = − 392 and − 21 x + 30 y = 3 . If we subtract the second equation from the first, we get 0 − 79 y = − 395 , so y = 5 . If we add the equations, we get − 42 x − 19 y = − 389 . So we want to multiply the second equation by 3.
Final Answer The first equation becomes − 21 x − 49 y = − 392 . We want to find k such that when we add the equations, the x terms are eliminated. So we want − 21 x + k ( − 7 x ) = 0 . This means − 21 − 7 k = 0 , so − 7 k = 21 , and k = − 3 . This is incorrect. We want the x term to be 21 x , so we need to multiply the second equation by 3. Then we have − 21 x − 49 y = − 392 and − 21 x + 30 y = 3 . If we subtract the second equation from the first, we get 0 − 79 y = − 395 , so y = 5 . If we add the equations, we get − 42 x − 19 y = − 389 . So we want to multiply the second equation by 3. If we multiply the second equation by 3, we get − 21 x + 30 y = 3 . Then we have − 21 x − 49 y = − 392 and − 21 x + 30 y = 3 . We want to eliminate the x terms, so we want to multiply the second equation by − 3 . Then we have 21 x − 30 y = − 3 . Adding this to − 21 x − 49 y = − 392 , we get − 79 y = − 395 , so y = 5 .
Conclusion The value of k is the factor by which the second equation should be multiplied. Therefore, the second equation should be multiplied by 3.
Examples
The elimination method is a fundamental technique in algebra used to solve systems of linear equations. It's not just theoretical; it has practical applications in various fields. For instance, consider a scenario where you're managing a budget and need to determine the cost of two different items. By setting up a system of equations and using the elimination method, you can efficiently find the individual costs, helping you make informed financial decisions. This method is also used in engineering to solve circuit problems and in economics to find equilibrium points in supply and demand models.
