A system of equations can be solved using linear combination to eliminate one of the variables.
[tex]
\begin{array}{l}
-y=-4 \quad \rightarrow \quad 10 x-5 y=-20 \\
x+5 y=59 \rightarrow \quad \frac{3 x+5 y=59}{13 x=39}
\end{array}
[/tex]
Which equation can replace [tex]$3 x+5 y=59$[/tex] in the original system and still produce the same solution?
A. [tex]$2 x-y=-4$[/tex]
B. [tex]$10 x-5 y=-20$[/tex]
C. [tex]$7 x=39$[/tex]
D. [tex]$13 x=39$[/tex]
Answer (2)
The equation that can replace 3 x + 5 y = 59 in the system and produce the same solution is D : 13 x = 39 , as it is satisfied by the derived values of x and y .
;
Solve the equation − y = − 4 to find y = 4 .
Solve the equation 13 x = 39 to find x = 3 .
Substitute x = 3 and y = 4 into each of the given equations.
The equation 13 x = 39 is satisfied by x = 3 , so it can replace 3 x + 5 y = 59 .
The final answer is 13 x = 39 .
Explanation
Understanding the Problem We are given a system of equations that is solved using linear combination. Our goal is to find an equation that can replace one of the original equations without changing the solution to the system. The original system is:
− y = − 4 3 x + 5 y = 59
which is transformed into:
10 x − 5 y = − 20 3 x + 5 y = 59
leading to
13 x = 39
Solving for x First, let's solve for x from the equation 13 x = 39 . Dividing both sides by 13, we get:
x = 13 39 = 3
Solving for y Next, let's solve for y from the equation − y = − 4 . Multiplying both sides by -1, we get:
y = 4
Checking the Equations Now we have the solution x = 3 and y = 4 . We need to find which of the given equations is satisfied by these values.
Let's test each equation:
2 x − y = − 4 :
2 ( 3 ) − 4 = 6 − 4 = 2 = − 4
So, this equation is not satisfied.
10 x − 5 y = − 20 :
10 ( 3 ) − 5 ( 4 ) = 30 − 20 = 10 = − 20
So, this equation is not satisfied.
7 x = 39 :
7 ( 3 ) = 21 = 39
So, this equation is not satisfied.
13 x = 39 :
13 ( 3 ) = 39
So, this equation is satisfied.
Final Answer Therefore, the equation that can replace 3 x + 5 y = 59 in the original system and still produce the same solution is 13 x = 39 .
Examples
Imagine you're baking a cake and need to adjust the ingredient ratios while keeping the final taste the same. This problem is similar; you're finding an equivalent equation to maintain the solution of a system, like adjusting flour and sugar amounts without changing the cake's flavor. Understanding how to manipulate equations while preserving their solutions is crucial in various fields, from engineering designs to financial modeling, where maintaining equilibrium under changing conditions is essential.
