# How do you solve simultaneous equations using the substitution method ?x=y 6x-2y=10 x=-y 3x-6y=36

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I am assuming that we are talking about 2 different systems since x=y in the first and x=-y in the second.

For the first system where x = y, simply substitute x for y. Then 6x - 2y becomes 6x -2x. So 6x -2x = 4x =10. Divide both sides by 4 and you have x = 2.5 => y = 2.5.

For the second system:

Since x = -y, then replace x with -y. Then 3x - 6y becomes 3(-y) - 6y. Now, we have -3y - 6y = -3y + -6y = -9y =36. Divide both sides by -9 and we have y= - 4 => -y = 4 => x = 4.

If you need to solve each with x=y and x=-y, then you would follow the same steps for the other equation. In other words, use x = -y for substituting in 6x - 2y. Then you would have 6(-y) - 2y = -6y - 2y = -8y = 10. Divide both sides by -8 and get y = -1.25. Since x= -y, the x = 1.25.

So these are two separate simultaneous equations, right? And x actually equals y in the first one?

If so, in the first one, just substitute y for x. Then you will have 6y - 2y = 10. That becomes 4y = 10 and that becomes y = 2.5.

The other one is a little more complicated since x = -y, but it's the same process. You substitute -y for x. Then you will have

-3y - 6y = 36

That gets you

-9y = 36

And y = -4

Since x = -y, x = -(-4). Two negatives make a positive and x = 4

Substitution Method is used to solve simultaneous equations, as given above. To solve simultaneous equations the following steps should be taken:

Label the equations as i and ii,

x=y ------(i)

6x-2y=10 ------(ii)

The next step is to isolate one variable from any equation, which has been done already as x=y.

After this step input the value of x in eq(ii) and solve to get the value of y:

6x-2y=10 ------(ii)

6y-2y=10

4y=10

4y/4=10/4 divide both sides by 4

**y=2.5**

After this we would have to input this value in equation ii, but since it has already been determined that x=y we can say that x=2.5 as well.

**x=2.5**

**y=2.5**

To further check the answers input both values in eq(ii)

6x-2y=10

6(2.5)-2(2.5)=10

15-5=10

10=10

LHS=RHS proved.

The same steps are to be followed in this one:

x=-y -------(i)

3x-6y=36 -------(ii)

Input the value of x in eq(ii)

3x-6y=36

3(-y)-6y=36

-3y-6y=36

-9y=36

-9y/-9=36/-9 divide both sides by -9

**y=-4**

Now since both value are equal only their signs are different i.e. x=-y therefore **x=4**.

Input both value in eq(ii) to check if they are correct:

3x-6y=36 -------(ii)

3(4)-6(-4)=36

12+24=36

36=36

LHS=RHS proved.

The substitution method involves replacing the variable in one equation with another variable. For example, in both cases we are told what x equals in terms of y. Replace every instance of x in the other equation with this y value. Then you will have only one equation in terms of y which you can solve for easily. Then plug this back into the other equation to solve for x.

There are two pairs pair of equations, each pair having two variables x and y.

First pair:

x=y.....................(i) and

6x-2y=10.............(ii)

are simultaneous linear equations in two variables x and y.

Since x=y , substitute y = x in (ii) and we get: 6x-2x = 10. Or 4x = 10 . Therefore, x = 10/4 = 2.5. Since y = x, y = 2.5

Second Pair:

x=-y ..................(1) and

3x-6y=36 ...........(2).

Though it is easy to substitute x= -y, we go different. From the second equation (dividing both sides by 3) we get x-2y = 12 . Or x = 12+2y. Substituting this x in terms of y in (1), we get:

12+2y = -y Or Subtracting 2y from both sides, we get:

12 = -3y Or y = 12/-3 = -4. So x =-y = -(-4) = 4.

Simultaneous equations are solved by substituting one variable for the other and setting the equation equal to 0, or if it has a constant. This is simply done by exchanging one for the other, for example, x=y, 6x-2y=10. You have to substitute y for x and the equation becomes:

6y-2y=10

4y=10

y=2.5

The other equations for y, substituting x for y is the same process.

3x-6y=36

-3y-6y=36

-9y=36

y=-4.

It is a very easy and straightforward process for you to solve similar equations using the same process.