Richard is 5 years older than shelia. three more than four times sheila age is 2 less than 3 times richards age . how old is each person

Richard is 15 and sheila ie 10. put the equations :r=s+5 and 4s+3=3r-2 and then solve

Let x be the age of Richard and y the age of Sheila."Richard is 5 years older than Shelia" can be represented by the equation x = y + 5.
"Three more than four times Sheila age is 2 less than 3 times Richards age":
Let's divide it into 2 parts: "Three more than four times Sheila" is 4y + 3, and "2 less than 3 times Richard's age" is 2 - 3x.
So "Three more than four times Sheila age is 2 less than 3 times Richards age" is 4y + 3 = 2 - 3x

We can now create a system:
x = y + 5 (1)
4y + 3 = 3x - 2 (2)

-Let's replace x by its algebraic value from equation (1) in equation (2):
4y + 3 = 3x - 2
4y + 3 = 3(y+5) - 2

-Develop 3(y+5):
3(y+5) = 3y + 15

4y + 3 = 3y + 15 - 2
4y + 3 = 3y + 13
Now subtract 3y and subtract 3 from both sides to have the variables on a side and the numbers on the other side:
4y + 3 - 3 - 3y = 3y - 3y + 13  - 3
y = 10

So Sheila is 10 years old.
And x = y + 5 = 10 + 5 = 15
So Richard is 15 years old.

You can recheck your
y + 5 = 10 + 5 = 15 = x (Richard is 5 years older than Sheila)
4y + 3 = 4*10 + 3 = 43 and 3x - 2 = 3*15 - 2 = 45 - 2 = 43 So 4y+3=3x-2=43 (Three more than four times Sheila age is 2 less than 3 times Richard's age)
The answer has been approved.

Hope this Helps! :)

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