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.
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.
8 answers
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Try better U can do it
4a= 3b
a=3 b=4
4(3)-7=C
C=5
The sides are 12, 12 ,and12
Curled from an internet Nd decided to share pls whosoever dat wrote it pls furgiv me
a=3 b=4
4(3)-7=C
C=5
The sides are 12, 12 ,and12
Curled from an internet Nd decided to share pls whosoever dat wrote it pls furgiv me
The solution is not well explained
I don't get it
I do not understand d solution pls explain.
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.
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.