SOLUTION: I am doing linear inequalities. I don't know how to set up the 2 equations to solve them. The perimeter of a rectangle is 20 inches. The length of the rectangle is 2 more than t

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Question 932897: I am doing linear inequalities. I don't know how to set up the 2 equations to solve them.
The perimeter of a rectangle is 20 inches. The length of the rectangle is 2 more than the width. Find the rectangles dimensions.
The perimeter of a triangle is 40 cm. The second side is 7 cm longer than the first and the third side is 9 units longer than the first. Find the lengths of each side.
You pull 17 coins out of a jar which contains dimes and pennies. The total value of the coins is $1.16. How many dimes and how many pennies do you have.
Thanks
Found 2 solutions by rothauserc, MathLover1:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
1) Perimeter(P) of rectangle = 2l + 2w where l is length and w is width
l = w+2
20 = 2(w+2) + 2w
20 = 2w +4 + 2w
4w = 16
w = 4 inches
l = 6 inches
2) let x be first side
P of triangle is the sum of its three sides
40 = x + x+7 + x+9
40 = 3x + 16
3x = 24
x = 8
first side is 8 cm
second side is 15 cm
third side is 17 cm
3) d + p = 17 where d is number of dimes, p is number of pennies
.10d + .01p = 1.16
solve first equation for p
p = 17 - d
now substitute for p in second equation
.10d + .01*( 17 - d) = 1.16
.10d + .17 - .01d = 1.16
.09d = .99
d = 11
p = 6
there are 11 dimes and 6 pennies


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
1.
The perimeter of a rectangle is 20in.
P=2L%2B2W=>
2L%2B2W=20in....eq.1
The length L of the rectangle is 2in more than the width W:
L=W%2B2in........eq.2
to find the rectangles dimensions,solve the system:
2L%2B2W=20in....eq.1
L=W%2B2in........eq.2..since this eq. already solved for L, substitute it in eq.1
__________________________
2%28W%2B2in%29%2B2W=20in....eq.1
2W%2B4in%2B2W=20in
4W=20in-4in
4W=16in
W=16in%2F4
highlight%28W=4in%29
now find L
L=W%2B2in........eq.2
L=4in%2B2in
highlight%28L=6in%29
answer: the rectangles dimensions are highlight%28L=6in%29 and highlight%28W=4in%29


2.
The perimeter of a triangle is 40cm.
P=a%2Bb%2Bc
a%2Bb%2Bc=40cm....eq.1
The second side b is 7cm longer than the first side a:
b=a%2B7cm....eq.2
and the third side c is 9 units longer than the first a:
c=a%2B9cm....eq.3

to find the lengths of each side, solve the system:
a%2Bb%2Bc=40cm....eq.1
b=a%2B7cm....eq.2->solved for b->substitute in eq.1
c=a%2B9cm....eq.3->solved for c->substitute in eq.1
_____________________
a%2B%28a%2B7cm%29%2B%28a%2B9cm%29=40cm....eq.1...solve for a
a%2Ba%2B7cm%2Ba%2B9cm=40cm
3a%2B16cm=40cm
3a=40cm-16cm
3a=24cm
a=24cm%2F3
highlight%28a=8cm%29
go to b=a%2B7cm....eq.2, substitute 8cm for a and solve for b
b=8cm%2B7cm
highlight%28b=15cm%29

go to c=a%2B9cm....eq.3, substitute 8cm for a and solve for c
c=8cm%2B9cm
highlight%28c=17cm%29

answer: the lengths of each side is highlight%28a=8cm%29 ,highlight%28b=15cm%29,and highlight%28c=17cm%29


3.
You pull 17 coins out of a jar which contains dimes=D and pennies=P. The total value of the coins is $1.16.

D%2BP=17...eq.1
let D be $0.10D and pennies be $0.01P
0.10D%2B0.01P=1.16...eq.2

solve the system:
D%2BP=17...eq.1
0.10D%2B0.01P=1.16...eq.2
______________________________
D%2BP=17...eq.1..solve for D
D=17-P...substitute in eq.2
0.10%2817-P%29%2B0.01P=1.16...eq.2
1.7-0.10P%2B0.01P=1.16
1.7-0.09P=1.16
1.7-1.16=0.09P
0.54=0.09P
0.54%2F0.09=P
highlight%28P=6%29
now find dimes
D=17-P
D=17-6
highlight%28D=11%29

answer: you have highlight%2811%29 dimes and highlight%28+6+%29 pennies