I need help on some elimination word problems.?

Question

- the difference of two numbers is 49. one half of the larger number plus one seventh of the smaller number is 56. find the numbers

- the sum of the length and width of a rectangle is 19 in. the length is one less than twice the width. find the length and width of the rectangle.

- the perimeter of a rectangle is 48 m. the width of the rectangle is 2 more than half the length. find the length and the width.

Answer 1

1)
let:
x=larger no.
y=smaller no.

x-y=49
x/2+y/7=56
------------------



x-y=49
14(x/2+y/7=56)14


~>

2(x-y=49)2
7x+2y=784


~>

2x-2y=98
7x+2y=784
-----------------
9x=882
9x/9=882/9
x=98(larger no.)


solve for y (use the 1st equation)

x-y=49
98(value of x we got)-y=49
-y=49-98
-y=-49

y=49(smaller no.)


2)

let

x=width of the rectangle
1-2x=length of the rectangle

x+(1-2x)=19
x+1-2x=19
-x=19-1
-x=18
x=-18

1-2x=1-2(-18)
=1-(-36)
=1+36
=37

3)

let
x=length
2+x/2=width


2(x)+2(2+x/2)=48
2x+4+x=48
3x=48-4
3x=44
3x/3=44/3

x=14.67

2+x/2=2+14.67/2
=2+7.33
=9.33

Answer 2

two numbers = x & y so 49 = x - y, they tell you this

x > y, they tell you this too

x/2 + y/7 = 56, this is also a given

Pick one to start with like "x" and define "x" using what you know:

if "x - y = 49" then "x = 49+y"

now go to the second equation and replace "x" with "49+7"
so in "x/2 + y/7 = 56" you now have " (49+y)/2 + y/7 = 56"

your bottom numers "2" and "7" are not the same so you need to make them the same by picking a number whose factors are 2 and 7. The easiest thing to do is just multiply then so 2 times 7 is fourteen.

Now, multiply the top numbers "(49+y) and "y" by the same amount you multiplied the number under it. So, since you multiplied 2 by 7 to make 14 you multiply the top, (49+y), by that to and you get 7 times (49+y) or 343 + 7y.

For the other, y/7, you multiplied 7 by 2 to get 14 so multiply the top, y, also by 2 and you get 2y.

Since you made the bottom numers the same you worry about the top numbers: "343+7y" and "2y". you can add the "y"'s together to get 9y and keep the regular number "343" so you get "343+9y" on top and "14" on the bottom.

Your new equation is (343 + 9y)/14 = 56
if you multiply both sides of the "=" by 14 you get:

"343+9y=56x14" or "343+9y=784"

if you subtract 343 from both sides you get "9y=784-343" or
9y=441

if you divide both sides by 9 you get y=441/9 or y=49

now you know y so go back to figure out x in the original equation of x - y = 49 but now you know y is 39 so you can write x-49=49.

if you add 39 to both sides you get x=49+49 which is x=98.

Now you have both numbers: x=98 and y=49
but you should check them by using the second equation they gave yo which was x/2+y/7=56 whichi gives you 98/2+49/7=56 or 49+7=56.

Answer 3

first question: the big number (x) is 98 and the small number (y) is 49

x-y=49 and (1/2)x+(1/7)y=56
x=49+y, so you can substitute 49+y for x
(1/2)(49+y)+(1/7)y=56
9y/14 + 49/2=56
9y/14=63/2
9y=441
y=49
and if you plug y back into x=49+y, 49+49=98, so x=98

second question: length is 37/3 in. and width is 20/3 in.

L+W=19 and L=2W-1
plug in 2W-1 for L
2W-1+W=19
3W=20
W=20/3, and if you plug W back in
L=2(20/3)-1=37/3

third question: length is 44/3 m. and width is 28/3 m.

2L+2W=48 and W=2+(1/2)L, so you can plug the second equation in for W
2L+2(2+(1/2)L)=2L+L+4=3L+4=48
3L=44
L=44/3
W=2+(1/2)(44/3)=28/3

Answer 4

(x - y = 49
(1/2x +1/7y = 56

x = 49 + y

(49 + y)/2 + y/7 = 56
LCD(2 e 7) = 14
343 + 7y + 2y = 784
9y = 784 - 343
9y = 441
y = 441 : 9
y = 49
x - y = 49
x - 49 = 49
x = 49 + 49
x = 98
Solution: (98, 49)
::
I) (2L + 2w = 19
II) (L = 2w - 1

2(2w - 1) + 2w = 19
4w - 2 + 2w = 19
6w = 19 + 2
w = 21 : 6
w = 3.5in
L = 2w - 1
L = 2(3.5) - 1
L = 7 - 1
L = 6in
Solution: L = 6in; w = 3.5in.
::
(2w + 2L = 48
(w = L/2 + 2 (-2)

(2w + 2L = 48
(-2w + L = -4
----------------
2w -2w + L + 2L = 44
3L = 44
L = 44/3
w = L/2 + 2
w = 44/6 + 2
w = 56/6 = 28/3
Solution: L = 44/3 and w = 28/3
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Answer 5

The first is 49 and 98.

Answer 6

it could be: 270 = 4x + 2(a hundred - x) a hundred heads is likewise, assuming that there at the instant are no substantial mutations, the style of persons. So, if there is x style of cows, the style of chickens is a hundred - x. Cows have 4 legs, and Chickens have 2. So, solved: 270 = 4x + 2(a hundred - x) 270 = 4x +200 - 2x 270 = 2x + 200 70 = 2x 35 = x a hundred - x = a hundred - (35) = sixty 5 There are 35 cows, and sixty 5 pigs.

Answer 7

the 3rd one is 20 and 4 i think

Answer 8

lol...i tried to work em out..but i only got a C in college algebra!! sorry! but i promise i tried several times!