Substitution Homework

Help with linear system substitution homework?

I'll do the first one because after you understand it, you'll understand how to do the 2nd one.

the linear substitution method is basically just isolating one of the variables.

1. x - 2y = -25
3x - y = 0

Isolate one of the variables in ONLY ONE OF THE EQUATIONS. I'm going to use the first equation and isolate x.

x = 2y - 25

Next you plug the x into the second equation to solve for y because that's the only variable left.

3(2y - 25) - y = 0
6y - 75 - y = 0
6y- y = 75
5y = 75
y = 15

Then you substitute the y back into one of the two equations (whichever one you prefer, because the answer for x will both be the same, which I'll prove)

x - 2 ( 15) = -25
x - 30 = -25
x = -25 + 30
x = 5

3x - 15 = 0
3x = 15
x = 15/3
x = 5

There you go. Hope you can get the rest on your own. Practice makes perfect!

Substitution method homework help?

expand equations
4x-8 = 8y-8 (second is already expanded)

get variables on left side of equals and constants on right side

4x-8y = 0 and 2x-16y = -36

use the first equation and solve for x
4x = 8y
divide by 4
x = 2y

now substitute this value for x into the second equation

2(2y) - 16y = -36
now solve for y
4y - 16y = -36
-12y = -36
y = 3

now plug this value for y into the first original equation and solve for x

4(x-2) = 8(3-1)
4x-8 = 16
4x = 24
x=6

so your answer as an ordered pair (x,y) is (6,3)

if you want to check your answer (always a good idea); plug into second original equation and it should be true

16(3)-32 = 2(6)+4
48-32 = 12+4
16 = 16

the answer is correct

hope this helps - M

The Substitution Method homework help please?

1.) c-d = 8
c/5 = d + 4
isolate "d" alone in the first equation.
d = c - 8
Now substitute it into the second equation.
c/5 = c-8+4
c = 5c - 40 + 20
4c = 20
c = 5
Substitute c = 5 back into either equation to solve for "d"
5 - d = 8
d = -3

2.) v = 9 - 2u
u/2 - 1/2 = -9 + 2u
u - 1 = -18 + 4u
3u = 17
u = 17/3

2(17/3) + v = 9
v = - 7/3

3.) y + x = 14
y = 14 - x
3(14-x) + 2x = 48
42 - 3x + 2x = 48
x = -6

3y + 2(-6) = 48
3y = 60
y = 20

4.) 3r + 2s = 6
3r = 6 - 2s
r = 2 - 2s/3
(2 - 2s/3)/4 + 2s/3 = -1
1/2 - 1s/6 + 2s/3 = -1
1s/2 = -3/2
s = -3

r/2 + (-3/3) = 1
r/2 = 2
r = 4

5.) x = 1000 - y
0.05(1000 - y) + 0.06y = 57
50 - 0.05y + 0.06y = 57
0.01y = 7
y = 700

x + 700 = 1000
x = 300

6.) x = 3y/2
3y/2 + y/2 = 10
3y + y = 20
4y = 20
y = 5

2x = 3(5)
2x = 15
x = 15/2

Hope this helps. And refer to the first example if you have troubles or msg me.

I need help with my math homework? It's substitution how do you do it?

I need help with my math homework? It's substitution how do you do it?
The directions say use substitution to solve each of the equations. If the system does not have exactly one solution, state whether it has no solution or infinetly many solutions.

1.) c-4d=1
2c-8d=2

2.) 2b=6a-14
3a-b=7

3.) x+y=16
2y= -2x+2

4.) x=2y
0.25x+0.5y=10

5.) x-2y= -5
x+2y= -1

6.) -0.2+y=0.5
0.4x+y=1.1

can you tell me how you got them as well. I need to learn this becasue we're haveing a test soon.

Marginal Rate of Substitution ? Homework problem, please help!?

Is it true that, in equilibrium, all consumers will have the same marginal rates of substitution
between purchased goods even though consumers differ in tastes? Why, or why not?

Substitution and Elimination math homework help!?

Substitution:
(1) 3x+y=4
(2) 5x-7y=11

Easiest eq to manipulate is (1), where y = 4 - 3x.
Substitute y into (2): 5x - 7(4-3x) = 11
5x - 28 + 21x = 11 Distribute -7 into parenthesis
26x - 28 = 11 Collect like terms
26x = 39 Add 28 to both sides
x = 3/2 Divide both sides by 26

Substitute x into the manipulated equation y = 4 - 3x
y = 4 - 3(3/2) = 4 - 9/2 = -1/2

Answer: (3/2, -1/2)



elimination:
(1) 3x+4y=7
(2) 2x-4y=13

y is easiest to eliminate because 4 is positive in (1) and negative in (2). Then adding (1) and (2),
5x = 20
x = 4
Substitute into either eq. (1) or (2). I'll do (1).
3(4) + 4y = 7
12 + 4y = 7
4y = -5
y = -5/4

Answer: (4, -5/4)



elimination:
(1) 2x+3y=1
(2) 5x+7y=3

Both equations have to be multiplied by a factor in order to eliminate. To eliminate x, multiply (1) by a factor of 5 (or -5) and (2) by a factor of -2 (or 2). To eliminate y, multiply (1) by a factor of 7 (or -7) and (2) by a factor of -3 (or 3). I'll eliminate x with 5 and -2.

10x + 15y = 5
-10x - 14y = -6
------------------------
y = -1

Substitute y into either equation. I'll use (1).
2x + 3(-1) = 1
2x - 3 = 1
2x = 4
x = 2

Answer: (2, -1)



word problem: assign x = # ppl ate chicken and y = #ppl ate beef (note: #ppl = #dishes served)
(1) x + y = 250
(2) 5x + 7y = 1500

To eliminate x, multiply (1) by - 5. Then add to (2).
-5x - 5y = -1250
5x + 7y = 1500
------------------------
2y = 250
y = 125

Using (1)
x + 125 = 250
x = 125

Answer: 125 dishes chicken and 125 dishes beef served

Freshman Algebra Homework. -Substitution Method.?

5
-x+9=6x-5
+x +x
9=7x-5
+5 +5
14=7x
x=2

y=-(2)+9
y=7

(2,7)

#6
5x+2(3x+7)=3
5x+6x+14=3
11x+14=3
-14 -14
11x=-11
x=-1

y=3(-1)+7
y=-3+7
y=4

(-1,4)

#7
x-y=11
+y +y
x=11+y

3(11+y)+10y=-6
33+3y+10y=-6
33+13y=-6
-33 -33
13y=-39
y=-3

x=11+(-3)
x=8

(8,-3)

#8
2x+2(4x+4)=13
2x+8x+8=13
10x+8=13
-8 -8
10x=5
x=1/2

y=4(1/2)+4
y=2+4
y=6

(1/2,6)

#9
5(y-1)-4y=-7
5y-5-4y=-7
y-5=-7
+5 +5
y=-2

x=-2+1
x=-1

(-1,-2)

I hope this helped! :D

Math homework question! Please help? Substitution or elimination?

c = cashews
1 - c = peanuts
This holds because we know that the mix is exactly 1 pound, and it won't be all cashews. Thus, whatever is not cashews (1 - c) will be peanuts.

2.80(1 - c) + 5.30(c) = 3.30
2.80 - 2.80c + 5.30c = 3.30
2.50c = 0.50
c = 0.2 = 20%
Thus, the mix is 20% cashews and 80% peanuts.
So, if you need to make 100 pounds, it will be 20 pounds of cashews and 80 pounds of peanuts!

I hope this helps! If it does, please feel free to select as best answer so I can get rewarded for my time and effort. I greatly appreciate it! Thanks and good luck!!!