Question

An equilatetal triangle with the lengths of its sides given in term of a and b.
If the lengths are 4acm,3bcm and (a+b+c)cm.
Find a and b and hence find the length of sides of the triangle.

Answer 1

the sides are equal, so

4a = 3b
4a = a+b+c

c = 3a-b, so

4(3a-b) = 3b
12a-4b = 3b
12a = 7b

subtracting,
8a = 4b
or
4a = 2b
But, 4a = 3b, so
2b = 3b
b=0

You only supplied two equations for three variables, and 0 is not a likely solution.

Check for typos and try again.

Answer 2

123%&&&5&&4%4427366&&*--3-%-$&-&--&%**&%

Answer 3

Try better U can do it

Answer 4

4a= 3b
a=3 b=4
4(3)-7=C
C=5
The sides are 12, 12 ,and12
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Answer 5

The solution is not well explained

Answer 6

I don't get it

Answer 7

I do not understand d solution pls explain.

Answer 8

From the question I think it should be (a+b+3)cm
4a=3b
4a=a+b+3
4a-3b=0
4a-a-b=3
=4a-3b=0_____(1) *3
3a-b=3_______(2) *4
Using elimination mathod
12a-9b=0
12a-4b=12
Minus
=-5b=-12
b=-12/-5
b=12/5
To find a substitute b in equation (1) to 12/5
At the end, (b=12/5 and a=9/5
Thank you.