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Thursday, October 19 2017 @ 04:39 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Algebra (No answer key)
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By: vivilau1112 (offline)  Sunday, June 23 2013 @ 08:55 AM CDT (Read 605 times)  
vivilau1112

Hello everyone! Can someone help me with these questions? Thanks in advance.

1.
(x-3)/5 = (y-7)/2

11x=13y

What is the value of x and y?


2.
3x+5 = 8y+4 = 7x-7y+1


What is the value of x and y?


3.
1.2x-0.8y=0.4 y+0.1x=0.3
What is the value of x and y?

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By: echeewh (offline)  Sunday, June 23 2013 @ 08:54 PM CDT  
echeewh

Hey there ,

Following pls find my worked solutions:

Q1.

(x-3)/5 = (y-7)/2
2(x-3) = 5(y-7)
2x - 6 = 5y - 35
5y - 2x = 35 - 6
5y - 2x = 29 -- (1)
11x - 13y = 0 -- (2)

Eliminate y,

(1)x13:
65y - 26x = 377 -- (3)

(2)x5:
55x - 65y = 0 -- (4)

(3)+(4):
55x - 26x = 377
29x = 377
x = 377 ÷ 29 = 13

y = 11x ÷ 13 = 11

==========

Q2.

3x + 5 = 8y + 4
8y + 4 = 7x - 7y + 1

Rearranging these 2 equations,

8y - 3x = 1 -- (1)

7x - 7y - 8y = 4 - 1
7x - 15y = 3 -- (2)

Eliminate x:
(1)x7:
56y - 21x = 7 -- (3)

(2)x3:
21x - 45y = 9 -- (4)


(3)+(4):
56y - 45y = 7 + 9
11y = 16
y = (16/11) = 1 5/11

in (1):
3x = 8y - 1 = [8 × (16/11)] - 1
= (128/11) - 1 = (117/11)
x = (117/11) ÷ 3 = (39/11) = 3 6/11

===========

Q3.

Multiply both equations by 10:

12x - 8y = 4 -- (1)
10y + x = 3 -- (2)

Eliminate x,

(2)x12:
120y + 12x = 36 -- (3)

(3)-(1):
120y + 8y = 36 - 4
128y = 32
y = 32 ÷ 128 = 0.25

in (1):
12x = 4 + 8y = 4 + (8 × 0.25)
= 4 + 2 = 6
x = 6 ÷ 12 = 0.5

===========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

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