the perimeter of a isosceles triangle is 28cm with side 4y=6y-2x+1 and base x+2y in cm, find x and y and state the length of the side of the triangle.
4y=6y-2x+1-> 2x-2y=1....eqn 1
4y+6y-2x+1+x+2y=28-> -x+12y=27....eqn2
multiply eqn2 by 2 to get
-2x+24y=54.......eqn3
eqn1+eqn2 gives
22y=55 which implies that y=2.5
substitute y=2.5 in eqn1 to get 2x-2(2.5)=1-> 2x=6 therefore x=3
Let
6y-2x+1=4y ——(1) for 2 equal length
And let
4y+6y-2x+1+x+2y=28 ——(2) for the perimeter of triangle
Now from equation (1)
6y-2x+1=4y
6y-4y=2x-1
2y=2x-1
y=(2x-1)/2
Also from equation (2)
4y+6y-2x+1+x+2y=28
12y-x=27 ——(3)
Substitute y in equation (3)
12((2x-1)/2))-x=27
6(2x-1)-x=27
12x-6-x=27
11x=33
x=33/11
x=3
Now substitute x in equation (3)
12y-x=27
12y-3=27
12y=27+3
12y=30
y=30/12
y=2.5
Therefore x=3 y=2.5 the length are a=8cm b=10cm c=10cm
Solved
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Thanks
I did not understand
It supposed to be
-2y=2x+1
Length of the triangle;
4y=4*2.5=10cm
6y-2x+1=6*2.5-2*3+1=10cm
X+2y=3+2*2.5=8cm
The length are 10cm,10cm,8cm
What abt the lengths of the triangle that should be stated
what do you mean
4y=6y-2x+1
?? Why not just say
2y = 2x-1
If that is the case, then we have
2(2x-1) + x+2x-1 = 28
Use that to find x, and then y.
If there's a typo in the problem, fix it and use the same logic.