SOLUTION: Hey can you please help me with this word problem ASAP? Five times my age augmented by twice my sisters age is equal to sixty-four. Likewise, triple her age decreased by twice my

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Question 1173233: Hey can you please help me with this word problem ASAP? Five times my age augmented by twice my sisters age is equal to sixty-four. Likewise, triple her age decreased by twice my age is equal to one. How old were each of us last year?
Found 3 solutions by Theo, ewatrrr, MathTherapy:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = your age.
let y = your sister's age.

5 times your age plus 2 times your sister's age = 64
the equation for that is 5x + 2y = 64

3 times her age minus 2 times your age = 1
the equation for that is 3y - 2x = 1

solve for x in the first equation as shown below.
start with 5x + 2y = 64.
subtract 2y from both sides of the equation to get:
5x = 64 - 2y
divide both sides of the equation by 5 to get:
x = (64 - 2y) / 5

replace x in the second equation with this to get:
3y - 2x = 1 becomes:
3y - 2 * (64 - 2y) / 5 = 1
multiply both sides of this equation by 5 to get:
15y - 2 * (64 - 2y) = 5
simplify to get:
15y - 128 + 4y = 5
combine like terms to get:
19y - 128 = 5
add 237 to both sides of this equation to get:
19y = 133
solve for y to get:
y = 133/19 = 7

replace y with 7 in the equation of 5x + 2y = 64 to get:
5x + 14 = 64
subtract 14 from both sides of the equation to get:
5x = 50
solve for x to get:
x = 50/5 = 10

you have x = 10 and y = 7

confirm by replacing x and y in both original equations to get:
5x + 2y = 64 becomes 5 * 10 + 2 * 7 = 64 which becomes 50 + 14 = 64 which becomes 64 = 64 which is true.
3y - 2x = 1 becomes 3 * 7 - 2 * 10 = 1 which becomes 21 - 20 = 1 which becomes 1 = 1 which is true.

you were 10 years old and she was 7 years old this year.

your solution is that you were 9 years old and she was 6 years old last year.


Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
2Equations, 2 unknowns
x your age
y sister's age
Question States:
5x + 2y = 64 Multiplying by 3 to eliminate y
-2x + 3y = 1 Multiplying by -2 to eliminate y
15x + 6y = 192
4x - 6y = -2
19x = 190
x = 10 , your age now, sister is -20 + 3y = 1, y = 21/3 = 7 years old now
How old were each of us last year? 9 and 6 years old a year ago
checking our Answers:
5x + 2y = 64
50 + 14 = 64 checks
Wish You the Best in your Studies.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Hey can you please help me with this word problem ASAP? Five times my age augmented by twice my sisters age is equal to sixty-four. Likewise, triple her age decreased by twice my age is equal to one. How old were each of us last year?
IGNORE Theo's setup of this problem. 
Never solve for a variable unless asked to use SUBSTITUTION or if one of the equations has variable coefficients, 1 or - 1.
Let my age and my sister's be I, and S, respectively
Then we get: 5I + 2S = 64 ------ eq (i)
Also, - 2I + 3S = 1 ------- eq (ii)
10I + 4S = 128 ------- Multiplying eq (i) by 2 ------ eq (iii)
- 10I + 15S = 5 ------ Multiplying eq (ii) by 5 ----- eq (iv)
19S = 133 ----- Adding eqs (iii) & (iv)
Sister's age, or highlight_green%28matrix%281%2C5%2C+S%2C+%22=%22%2C+133%2F19%2C+%22=%22%2C+7%29%29
5I + 2(7) = 64 ------ Substituting 7 for S in eq (i)
5I + 14 = 64
5I = 50
My age, or highlight_green%28matrix%281%2C5%2C+I%2C+%22=%22%2C+50%2F10%2C+%22=%22%2C+10%29%29
I'll now leave you to find both ages, a year ago!