9th Grade math questions; Best answer quickly, if you know these

These are simultaneous equations.

1) In three years time a pet mouse will be as old as his owner was four years ago. Their present ages total 13 years. Find the age of each now.

2) If the numerator and denominator of a fraction are both decreased by one the fraction becomes 2/3. If the numerator and denominator are both increased by one the fraction becomes 3/4. Find the original fraction.

3)The curve y=ax^2+bx+c passes through the points (1,8), (0,5) and (3,20). Find the values of a, b and c and hence the equation of the curve.

You dont need to answer all of them, if you don't know... And not a lot of explanation is required just couple of steps =)

Thanks is advance! ;)

1) Let's say that the mouse is now "x" years old, and that the owner is "y" years old.

"Their present ages total 13 years."
x + y = 13
y = 13 - x … eqn 1

"In three years time a pet mouse will be as old as his owner was four years ago."
x + 3 = y - 4 … eqn 2

Substitute eqn 1 into eqn 2:
x + 3 = (13 - x) - 4
x + 3 = 9 - x
x + x = 9 - 3
2x = 6
x = 3

From eqn 1, y = 10

SOLUTION: The mouse is now 3 years old, and the owner is 10 years old.

(Check: in three years' time, the mouse will be 6. Four years ago, the owner was 6.)

––––––––––––––––––––––––––––

First set: Nadal 7-5

2) Let's call the original fraction x/y

"If the numerator and denominator of a fraction are both decreased by one the fraction becomes 2/3."
(x - 1)/(y - 1) = 2/3

Cross multiply:
3(x - 1) = 2(y - 1)
3x - 3 = 2y - 2
3x - 1 = 2y … eqn 1

"If the numerator and denominator are both increased by one the fraction becomes 3/4."
(x + 1)/(y + 1) = 3/4

Cross multiply:
4(x + 1) = 3(y + 1)
4x + 4 = 3y + 3
4x + 1 = 3y … eqn 2

Using the elimination method … 3(eqn 1) – 2(eqn 2):
9x - 3 = 6y –
8x + 2 = 6y
–––––––––
x - 5 = 0
x = 5

Plug this "x" value into either original equation, in this case eqn 1:
3(5) - 1 = 2y
15 - 1 = 2y
14 = 2y
y = 7

SOLUTION: The original fraction is 5/7

(Check: 4/6 = 2/3, and 6/8 = 3/4)

––––––––––––––––––––––––––––

(Back after the second set in the tennis.)

First two questions were answered correclty by Jacinta. So the third one:
y=ax^2+bx+c passes through (1,8), (0,5), (3,20)
In the Eqn., put x=1, y=8,
8=a+b+c - - - >Eqn. 1
In the Eqn., put x=0, y=5,
5=0+0+c
5=c - - - > Eqn. 2
In the Eqn., put x=3, y=20,
20=9a+3b+c - - - > Eqn. 3
Solving Eqn. 1, 2 and 3
a=1, b=2, c=5