Question

5. Here are clues for a puzzle involving two numbers: Seven times the firstnumber plus six times the second number equals 31. Three times the firstnumber minus ten times the second number is 29. What are the twonumbers?

Answer 1

$\begin{gathered} \text{first number=}\frac{11}{2} \\ \text{second number=}\frac{-5}{4} \end{gathered}$

Explanation

Step 1

let

x represents the first number

y represents the second number

Hence,

a)Seven times the first number plus six times the second number equals 31

so, traslating

$7x+6y=31\rightarrow equation(1)$

b)Three times the first number minus ten times the second number is 29

$3x-10y=29\rightarrow equation(2)$

Step 2

now. we have a system of equations, Let's solve it !

$\begin{gathered} 7x+6y=31\rightarrow equation(1) \\ 3x-10y=29\rightarrow equation(2) \end{gathered}$

a) isolate x in equation (1) and replace in equation(2)

$\begin{gathered} 7x+6y=31\rightarrow equation(1) \\ \text{subtract 6y in both sides} \\ 7x+6y-6y=31-6y \\ 7x=31-6y \\ \text{divide both sides by 7} \\ \frac{7x}{7}=\frac{31-6y}{7} \\ x=\frac{31}{7}-\frac{6}{7}y\rightarrow equation\text{ (3)} \\ \text{now , replace eq(3) into eq(2)} \\ 3x-10y=29\rightarrow equation(2) \\ 3(\frac{31}{7}-\frac{6}{7}y)-10y=29 \\ \frac{93}{7}-\frac{18}{7}y-10y=29 \\ \frac{93}{7}-\frac{88}{7}y=29 \\ \text{subtract 93/7 in both sides} \\ \frac{93}{7}-\frac{88}{7}y-\frac{93}{7}=29-\frac{93}{7} \\ -\frac{88}{7}y=\frac{110}{7} \\ -88y=110 \\ y=\frac{110}{-88} \\ y=\frac{55}{44}=-\frac{5}{4} \\ so \\ x=-\frac{5}{4} \end{gathered}$

b) now, replace the y value into eq(3)

$\begin{gathered} x=\frac{31}{7}-\frac{6}{7}y\rightarrow equation\text{ (3)} \\ x=\frac{31}{7}-\frac{6}{7}(-\frac{5}{4}) \\ x=\frac{31}{7}+\frac{30}{28} \\ x=\frac{11}{2} \end{gathered}$

therefore , the answer is

$\begin{gathered} \text{first number=}\frac{11}{2} \\ \text{second number=}\frac{-5}{4} \end{gathered}$

I hope this helps you