Solve the system for $x$ and $y$.
$\begin{array}{l}
7 x+8 y=26 \\
-12 x-7 y=-58
\end{array}$
A. $(14,-9)$
B. $(-2.5)$
C. $(6,-2)$
D. $(-1.10)$
Answer (2)
The solution to the given system of equations is ( 6 , − 2 ) . This was determined by using the elimination method to eliminate one variable and solve for the other. After substitution back into the original equation, both values were confirmed to be correct.
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Multiply the first equation by 12 and the second equation by 7 to prepare for eliminating x .
Add the modified equations to eliminate x and solve for y , resulting in y = − 2 .
Substitute y = − 2 into the first original equation and solve for x , resulting in x = 6 .
The solution to the system of equations is ( 6 , − 2 ) .
Explanation
Analyze the problem We are given a system of two linear equations with two variables, x and y . Our goal is to find the values of x and y that satisfy both equations simultaneously. The given equations are:
7 x + 8 y = 26 − 12 x − 7 y = − 58
We can solve this system using either the substitution or elimination method. Here, we will use the elimination method.
Eliminate x To eliminate x , we can multiply the first equation by 12 and the second equation by 7. This gives us:
12 ( 7 x + 8 y ) = 12 ( 26 ) 7 ( − 12 x − 7 y ) = 7 ( − 58 )
Which simplifies to:
84 x + 96 y = 312 − 84 x − 49 y = − 406
Solve for y Now, we add the two equations to eliminate x :
( 84 x + 96 y ) + ( − 84 x − 49 y ) = 312 + ( − 406 )
This simplifies to:
47 y = − 94
Calculate y Dividing both sides by 47, we get:
y = 47 − 94 = − 2
Substitute y into the first equation Now that we have the value of y , we can substitute it back into either of the original equations to solve for x . Let's use the first equation:
7 x + 8 y = 26
Substitute y = − 2 :
7 x + 8 ( − 2 ) = 26 7 x − 16 = 26
Isolate x Add 16 to both sides:
7 x = 26 + 16 7 x = 42
Calculate x Divide both sides by 7:
x = 7 42 = 6
State the solution Therefore, the solution to the system of equations is x = 6 and y = − 2 . So the solution is ( 6 , − 2 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, suppose a company wants to know how many products they need to sell to cover their costs. They can set up a system of equations where one equation represents the cost of production and the other represents the revenue from sales. Solving this system will give them the number of products they need to sell to break even. In this case, the solution (6, -2) represents a specific point where two linear relationships intersect, providing valuable information for decision-making.
