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 answer: 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.
