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=}(11)/(2) \\ \text{second number=}(-5)/(4) \end{gathered}`

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

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

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

therefore , the answer is

`\begin{gathered} \text{first number=}(11)/(2) \\ \text{second number=}(-5)/(4) \end{gathered}`

I hope this helps you