two questions of maths (F.2) 15 points!, emergency!

Ok. Let me answer your question=)

QUESTION ONE

5ax-2ay = 7x
a (5x-2y) = 7x -------------------(抽個a出黎，因為a係公因子common factor)

a = 7x / (5x-2y) ------------- 等號兩邊一齊除以(5x-2y)

當x=3, y=1,

a = 7x / (5x-2y) = 7(3) / [ 5(3)-2(1)] = 21 / (15-2) = 21/13

QUESTION TWO
Let the length be y (cm)
Then the width of the rectangle is y-6 (cm)

Hence the area of the rectangle is: y (y-6) = 135

y^2 - 6y - 135 = 0
(y-15)(y+9) = 0 (因式分解 factorization)

Hence (y-15) = 0 OR (y-9)=0

Hence y = 15 or 9

***CHECKING

if y = 9, length = 9cm and width = 3cm, the area is 9x3 = 27cm^2 (WRONG)
HENCE y = 9 is NOT A ROOT

if y = 15, length = 15cm and width = 9cm, the area is 15x9 = 135cm^2 (CORRECT)

Hence y=15 is a correct ROOT

ANSWER:
LENGTH = 15cm

我希望幫到你=)

1.
Given 5ax-2ay=7x, express a in terms of x and y. Find the value of a when x=3 and y=1

5ax-2ay=7x

a(5x - 2y) = 7x

a = 7x/(5x - 2y)

when x = 3 and y = 1

a = 21/(15 - 2)

a = 21/13


2.the length of an rectangle is 6cm longer than its width. If its area is 135cm square,

find its length

let x cm be the length,then the width is x - 6 cm

x(x - 6) = 135

x^2 - 6x - 135 = 0

(x - 15)(x + 9) = 0

x = 15 or x = -9(rejected)

so,its length is 15cm