Solve ea. of the following systems by substitution.?

Question

Solve ea. of the following systems by substitution. Can some please help me and show me how to work? Thank you.

1. 5x – 2y = -5
y – 5x = 3

2. 8x – 4y = 16
y = 2x – 10

3. 4x – 12y = 5
-x + 3y = -1

Answer 1

(1)
5x – 2y = -5.........1
y – 5x = 3..........2
by(2)eqe
y=5x+3
put the value of y in (1)then
x=-1/5
y=2

result of 2 and 3 question is not possible with this mathod.

Answer 2

1. 5x – 2y = -5
y – 5x = 3 ----> y = 3 + 5x
5x – 2(3 + 5x) = -5
5x - 6 + 10x = -5
15x -6 = -5
15x = 1
x = 1/15
y = 3 + 5(1/15) = 3 + 3 = 6

2. 8x – 4y = 16 --->
y = 2x – 10
8x – 4(2x – 10) = 16
8x - 8x + 40 = 16
40 = 16 not true, so this cannot be solved using substitution (I also tried to solve this with elimination.) There is not a solution set that will work for this system of equations.

3. 4x – 12y = 5
-x + 3y = -1 ----> x = 3y + 1
4(3y + 1) – 12y = 5
12y + 4 - 12y = 5
4 = 5 not true - This is the same situation as above.

Answer 3

Substitution means that you solve one of the equations for a variable and then *substitute* the answer onto the other equation.

1. 5x - 2y = -5
y - 5x = 3 ---> We're going to use this one because one of the variables doesn't have a coefficient.

y = 5x +3 ---> Add 5x to both sides.

Now we can substitute (5x + 3) for y in the first equation.

5x - 2(5x+3) = -5
5x -10x - 6 = -5 ---> Distributive property.
-5x - 6 = -5 ---> Add like terms.
-5x = 1 ---> Add 6 to both sides.
x = -1/5 ---> Divide both sides by (-5) and we have an answer for x.

Insert (-1/5) for x in the second equation to solve for y.

y - 5(-1/5) = 3
y +1 = 3 ---> Multiply.
y = 2 ---> Subtract 1 from both sides.

The answer is (-1/5, 2).

2. 8x - 4y = 16
y = 2x - 10 ----> Since y has already been solved for, just insert it into the first equation.

8x - 4(2x - 10) = 16
8x - 8x + 40 = 16 ---> Distributive property.
40 = 16 ---> Since the numbers are not the same, this equation has no solution.

3. 4x - 12y = 5
-x + 3y = -1 ---> I'll use this equation because the x has almost no coefficient.

-x + 3y = -1
-x = -3y -1 ---> Subtract 3y to both sides.
x = 3y +1 ---> Multiply both sides by -1.

Insert (3y + 1) for x in the first equation.

4(3y + 1) - 12y = 5
12y +4 - 12y = 5 ---> Distributive property.
4 = 5 ---> Add like terms. These equations are like problem #2, because they have no solution.

Answer 4

Hello associates07... weird name...

Well it could be that you are trying to cheat in your math homework? It will not make you get ahead in the long run. However here are the answers..cos i was stuck in math and i cheated too!!

1) You can multiply the top equation by a number to make all the values increase or reduce in value to match the lower equation. In this case you don't need to.

5x - 2y = -5
y - 5x = 3 {add both equations together to give you...}
^^^^^^^^^^^^^^^
-y = -2 {this is because when you add (-5x) to (+5x) it is 0}

Since y = 2

Substitute back into equation:

5x - (2 * 2) = -5 {use the known value of 'y' in the equation}
5x - 4 = -5
5x = -5 + 4 {it becomes positive on the other side}
5x = -1

x = (-1/5 in fraction) or (-0.2 decimal)

therefore x = -0.2 and y = 2

2) Again the same principal can be used; adjust the equation to something easy for you to use! Take control and manipulate!

y = 2x - 10

is exactly the same as

y - 2x = -10

Remember I mentioned earlier about having to multiply the equation to make it workable?

now; (y - 2x = -10) * 4 = 4y - 8x = -40

adding this to the other original equation -
8x - 4y = 16
4y - 8x = -40
0 = -24
{We have shown that the simultaneous equation was not correct/workable, because the nought cannot equal minus 24! Or it could be that you have done a typo! If so try it yourself, and i'll post the answer only....NEXT!}

3) Again the solution is not achievable, but you can prove it by the previous method.

Answer 5

1. 5x – 2y = -5
y – 5x = 3

To solve by substitution, first place one of the equations in terms of one of the variables. I'll solve the second equation for y, since it has no coefficient:

y - 5x = 3
y = 3 + 5x

Then substitute (replace "y" with "3+5x") in the other equation:

5x – 2y = -5
5x - 2(3 + 5x) = -5
5x - 6 - 10x = -5
5x = -1
x = -1/5

Then go back and use either equation, with the now-known value for x, and solve for y:

y - 5x = 3
y - 5(-1/5) = 3
y + 1 = 3
y = 2

2.
8x - 4y = 16
y = 2x - 10

8x - 4y = 16
8x - 4(2x - 10) = 16
8x - 8x + 40 = 16
40 = 16

When you get a result like this -- the variables all cancel and you are left with a false equality -- it indicates that are no solutions. These two equations define parallel lines that never intersect.

3.
4x - 12y = 5
-x + 3y = -1

-x + 3y = -1
x = 3y + 1

4x - 12y = 5
4(3y + 1) - 12y = 5
12y + 4 - 12y = 5
4 = 5

Same as #2.

Answer 6

1. 5x - 2y= -5
y-5x=3

You want to make the second one equal to y. So you get y=5x+3 and you plug that into the first equation so you get...

5x - 2 (5x+3)= -5
5x - 10x - 6 = -6
-5x = 0
x=0

2. 8x - 4y =16
y=2x-10

8x - 4 (2x -10)=16
8x - 8x +40 = 16
0x= - 24
This one cant be solved because you cant divide by 0.

3.4x - 12y=5
-x + 3y = -1
x=3y+1

4(3y+1) - 12y=5
12y +4 -12y=5
0y=-1

same thing with this one.

Hope this helps.

Answer 7

1) 5x - 2y=-5 .......(i)
-5x + y=3 ........(ii)
Add equations (i) and (ii) togther to have;
5x-5x-2y+y=-5+3
-y=-2; y=2.
by putting the value of y into,say equation(ii)
-5x+2=3
-5x=1; x=-1/5

2) 8x-4y=16 .......(i)
y=2x-10 .......(ii)
equation(i) can be simplified, by dividing all through by 4;
2x-y=4 ......(i)
-2x+y=-10......(ii)
from the above equation, it can be clearly seen that the value of x and y cannot be found when adding both equations together.Therefore x and y assumes no value for the above equation.Hope it's well understood?

3) 4x-12y=5.....(i)
-x+3y=-1 ......(ii)
we can multiply eq.(ii) all through by 4,in order that, adding both eq.(i)and(ii) will nullify 'x' from the equation(i.e we'll have sth like 4x-4x) so as to get the value for y.
4x-12y=5...(i)
-4x+12y=-1..(ii)
addig both equations seems to nullify both the x and y value. Hence x and y have no value for the above equation.Just as the previous example we've just solved.
Thanks.

Answer 8

1.5x – 2y = -5 _______(i)
y – 5x = 3 ________(ii)


(sol.) consider the eqn. y-5x=3
let y=3+5x
put y=3+5x in 5x-2y=-5 as
5x-2(3+5x)= -5 simpfying it as
5x-6-10x= -5
now
5x-10x-6+6= -5+6 adding 6 to bothsides we get
-5x= 1
-5x/-5= 1/-5 dividing bothsides by 5 we get
x= -1/5 is the answer
now put x= -1/5 in 5x-2y= -5 as
5(-1/5)-2y= -5
now after simplify[ng we have
-1-2y=-5
-1+1-2y= -5+1 adding 1 to bothsides we get
-2y= -4 dividing by 2 to bothsides we get
-2y /2= -4/2
-y= -2 multiplying -1 to bothsides we get
y=2 is answer



2.8x – 4y = 16 ________(i)
y = 2x – 10 ________(ii)

sol.
put y=2x-10 in eqn. 8x-4y=16 as
8x-4(2x-10)=16
by simplifying the left handside we get
8x-8x-40=16
now we have -40=16 solution is not possible


3.4x – 12y = 5 ______(i)
-x +3y = -1 ______(ii)
sol.
consider the eqn.-x+3y= -1
now by subtracting 3y to bothsides we get
-x+3y-3y= -1-3y
now we have -x= -1-3y
multiply above eqn. by -1 we get
x=1+3y
now put x=1+3y in eqn(i) as
4(1+3y )-12y=5
now after simplifying we get
4+12y-12y=5
i-e 4=5 hence solution of this eqnuation is not possible

Answer 9

in each question their asking you to solve one of the equation for x or y and then substitute that into the other equation.

1) 5x-2y=-5
y-5x=3

so solve one of the equation for x or y, the second equation is fairly simple to solve for y,

y-5x=3

==> add 5x to both sides

y=3+5x

now wherever you see "y" in the first equation replace it with "3+5x"

5x-2(3+5x)=-5

now simplify and solve for x

5x-6-10x=-5

==> combine all x terms

-5x-6=-5

==> add 6 to both sides

-5x=1

divide through by -5

x=-1/5

now substitute -1/5 into y=3+5x to get y

y=3+5(-1/5)

y=3-1

y=2

so x is -1/5 and y=2

Answer 10

1.)5x -2y= -5
-2y+10x= -6 mult.by -2
addiing 1 and 2
15x=-10
so,x= -2/3 and y= 5/6

2.) 8x-4y=16
8x-4y=40 mult. by 4
well ,this is not possible


3.)4x-12y=5
4x-12y=4 mult.by-4
this is also not possible