Question

a larger number is 7 less than 5 times a smaller number. the sum of the two numbers is 41. find the numbers

Answer 1

Answer:

Step-by-step explanation:

Let's call the first number x and the second y.

From the first condition $x+7 = 5y$; while from the second $x+y=41$

It's a simple system of equation, I'd recommend solving by solving both for x and comparing the results.

$\left \{ {{x=5y-7} \atop {x=41-y}} \right.\\5y-7 = 41-y \rightarrow 6y=48 \rightarrow y =8;\\x= 41 - 8 \rightarrow x = 33$

Answer 2

Answer:

(8,33) or x=8 and y=33

Step-by-step explanation:

A number, x or y is less than 5 times a smaller number: y large number and x smaller number

y=5x-7

y+x=41 is also y=-x+41

5x-7=-x+41

5x=-x+48

6x=48

x=8

Plug in 8=x to the equation

y+8=41

y=33