Sum of three times a number and four times another number is 68. Subtracting two times the second number from four times the first number gives 10. What are the numbers?

The two numbers are 8 and 11. The first number is represented by x and the second by y, and both satisfy the given equations. The solution was verified by substituting back into the original equations.
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Answered by Anonymous | 2025-07-04

To solve this problem, we need to set up a system of equations based on the information given. Let's break it down step by step.

Define the Variables:

Let x be the first number.
Let y be the second number.


Set Up the Equations:

The first part of the problem states, "Sum of three times a number and four times another number is 68." This translates to the equation:
3 x + 4 y = 68

The second part says, "Subtracting two times the second number from four times the first number gives 10." This gives us the second equation:
4 x − 2 y = 10



Solve the System of Equations: We will use the method of elimination or substitution to solve these equations.
Using Elimination:

First, we can multiply the second equation by 2 to align the coefficients of y :
8 x − 4 y = 20

Now, we add this equation to the first equation to eliminate y :
[\begin{align*}




3x + 4y &= 68 \ 8x - 4y &= 20 \ \hline 11x &= 88 \ \end{align*}]
* Divide both sides by 11 to solve for [tex]x[/tex]:

\[x = \frac{88}{11} = 8\]

* Substitute [tex]x = 8[/tex] back into the first equation to find [tex]y[/tex]:

\[3(8) + 4y = 68\]
\[24 + 4y = 68\]
\[4y = 44\]
\[y = \frac{44}{4} = 11\]



Conclusion:

The first number x is 8.
The second number y is 11.



These numbers satisfy both conditions of the problem.

Answered by DanielJosephParker | 2025-07-07