SOLUTION: system of Equations and inequalities There are 2 triangles which I can't draw, imagine them between the letters and numbers. Could someone please help with this? Could you show al

Question 554827: system of Equations and inequalities
There are 2 triangles which I can't draw, imagine them between the letters and numbers. Could someone please help with this? Could you show all work and explain steps for me so I understand.
The triangle on the left has a perimeter of 14. The triangle on the right has a perimeter of 21. What are x and y?

1st triangle has a y on each side and x at bottom the other triangle has 5/4y on each side and 3x at the bottom.

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The triangle on the left has a perimeter of 14. The triangle on the right has a perimeter of 21. What are x and y?
1st triangle has a y on each side and x at bottom the other triangle has 5/4y on each side and 3x at the bottom.
:
The way I understand this, the two triangles have two equal sides and writing a perimeter equation for each, I have:
:
Left triangle: x + 2y = 14
Right triangle: 3x + 2()y = 21
On the right triangle, let's get rid of the fraction, multiply each term by 4
4(3x) + 2(5y) = 4(21)
12x + 10y = 84
;
Back to the left triangle
x + 2y = 14
x = (-2y + 14); put it in this form for substitution
:
Back to the right triangle
12x + 10y = 84
replace x with (-2y+14), find y
12(-2y+14) + 10y = 84
-24y + 168 + 10y = 84
-24y + 10y = 84 - 168
-14y = -84
Mult by -1
14y = 84
y = 84/14
y = 6
Find x
x = -2(6) + 14
x = -12 + 14
x = 2
Summarize solutions, x=2, y=6
:
Check this in the left equation
x + 2y = 14
2 + 2(6) = 14
also in the right equation
3x + 2()y = 21
3(2) + 2()6 = 21
mult 5 by 6
6 + 2() = 21
Cancel the 2 into the 4
6 + =
6 + 15 = 21; confirms our solutions on this one too
:
did I succeed in explaining the step to you here? Any questions? C
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