Algebra Solutions

x + 3 = 9 (original equation )
x = 9 – 3 (take the constant term to the other side so that the variable x is isolated.)
x = 6

So x = 6 is the required solution to the equation.

3x = 9 (original equation)
x = 9/3 (take the constant term associated with x to the other side so that the variable x is isolated.)
x = 3

So x = 3 is the required solution to the equation.

In solving these equations, we use a simple Algebraic technique called "Substitution Method". In this method, we evaluate one of the variable value in terms of the other variable using one of the two equations. And that value is put into the second equation to solve for the two unknown values.

The solution below will make the idea of Substitution clear.

Using 1^st equation,

x – y = 10 -----(1)
x = 10 + y

Now we put this value of x into the 2^nd equation.

x + y = 15 -----(2)
(10 + y) + y = 15
10 + 2y = 15
2y = 15 – 10 = 5
y = 5/2

Putting this value of y into any of the two equations will give us the value of x.

x + y = 15
x + 5/2 = 15
x = 15 – 5/2
x = 25/2

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.

Elimination Method.

In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.

In this example, we see that the coefficients of all the variable are same, i.e., 1. So if we add the two equations, the –y and the +y will cancel each other giving as an equation in only x. Let me illustrate this below.

x – y = 10
x + y = 15
2x = 25
x = 25/2

Putting the value of x into any of the two equations will give y = 5/2

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.

Elimination Method - By Equating Coefficients:

In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.

This is another very easy and useful equation solving technique that is extensively used in Algebraic calculations. We illustrate this method through an example.

In this example, we see that neither the coefficients of x nor those of y are equal in the two equations. So simple addition and subtraction will not lead to a simplified equation in only one variable. However, we can multiply a whole equation with a coefficient (say we multiply equation (2) with 2) to equate the coefficients of either of the two variables.

After multiplication, we get

2x + 4y = 30 ------(2)'

Next we subtract this equation (2)’ from equation (1)

2x – y = 10
2x + 4y = 30
–5y = –20
y = 4

Putting this value of y into equation (1) will give us the correct value of x.

2x – y = 10 ------(1)
2x – 4 = 10
2x = 10 + 4 = 14
x = 14/2 = 7

Hence (x , y) =( 7, 4) gives the complete solution to these two equations.

We are given that

4x – 3 = 3x + 8

Separating the variables and the coefficients gives:

4x – 3x = 8 + 3

(Note: Taking a constant or a variable term to the left hand side from the right hand side (or vice versa) changes its sign as illustrated above.)

Simplifying the above equation on the L.H.S (Left Hand Side) and the R.H.S (Right Hand Side) gives

x = 11

Hence x = 11 is the required solution to the above equation.

In the equation Ax + B = Cx + D, the coefficients A, B, C, D may also be any decimal numbers. For example, the equation could be of this form: 4x + 3.2 = 6.1x + 5.2 -- But you are not supposed to be confused with the method. There will be no change in the equation solving strategy and once you have learnt the above method, you do not need to bother about the coefficients at all.

Next we present and try to solve the examples in a more detailed step-by-step approach. Examples given next are similar to those presented above and have been shown in a way that is more understandable for kids.

We are given that

x + 4 = 11

(Subtracting 4 from both sides of the equation gives)

x + 4 – 4 = 11 – 4
x = 7

Hence x = 7 is the solution to the given equation.

If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.

2x + y = 15 ------(1)
3x – y = 10 ------(2)
______________

5x = 25

(Since +y and –y cancel out each other)

What we are left with is a simplified equation in x alone. i.e., 5x = 25

(Dividing this equation throughout by 5 gives)

5x/5 = 25/5
x = 5

(Putting this value of x into equation (1) gives)

2(5) + y = 15
10 + y = 15

Which is another equation in a single variable y.

(Subtracting 10 from both sides of the equation gives)

10 + y – 10 = 15 – 10
y = 5

Hence the solution to the system of equations is (x , y) = (5, 5)

With a little observation, we can conclude that if we directly add these two equations, we are not going to reach any simple equation. Let us show this below.

x + 2y = 15 ------(1)
x – y = 10 ------(2)
______________

2x + y = 25

Which is another equation in 2 variables x and y. So our problem doesn’t seem to have reduced. If instead of adding the two equations directly, I multiply the entire equation (1) with – 1, and then add the resulting equation into equation (2), the +x will be cancelled out with – x as shown next.

(Multiplying equation (1) with – 1 on both sides of the equality gives)

– ( x + 2y ) = – 15
– x – 2y = – 15 ------(1’)

(Adding the new equation (1’) to the equation (2) gives)

– x – 2y = – 15 ------(1’)
x – y = + 10 ------(2)
______________
– 3y = – 5

(Dividing on both sides of the equation by – 3)

-3y/-3=-5/-3
y = 5/3

(Putting this value of y into equation (2) gives)

x – 5/3 = + 10

(Adding 5/3 to both sides of the equation gives)

x – 5/3 + 5/3 = 10 + 5/3
x = (30+ 5)/3 = 35/3

Hence the solution to the given system of equations is (x , y) = ( 35/3 , 5/( 3 ))

Apparently, this system seems to be a bit complex and one might think that no cancellation of terms is possible. But a close observation and a simple multiplication can lead us in the right direction.

We are given two equations:

8x – 13y = 2 ------(1)
–4x + 6.5y = –2 ------(2)

(Multiplying the entire equation (2) by 2 gives)

2(–4x + 6.5y ) = 2(–2)
–8x + 13y = –4 ------(2’)

(Adding the new equation (2’) to the equation (1) gives)

8x – 13y = 2 ------(1)
–8x + 13y = –4 ------(2’)
______________
0 = –2

But this is not true!! 0≠ –2

Hence the two equations constitute an inconsistent system of linear equations and thus do no have a solution (At no point do the two straight lines intersect => No solution!)

In this method of equation solving, we work out on any of the given equations for one variable value, and then substitute that value in the other equation. It gives us an equation in a single variable and we can use a single variable equation solving technique to find the value of that variable (as shown in examples above). Let us solve the given system now

We are given that

2x – 2y = –2 ------(1)
x + y = 24 ------(2)

Now the next question is: which equation to pick up. There is no particular criteria for this choice. One can simply choose an equation that makes the calculations simpler. E.g., in this example, the equation (2) is easier to work on.

x + y = 24 ------(2)

(we find the value of x in terms of y)

x = 24 –y

(Next we put this value of x into equation (1))

2(24 – y) – 2y = –2
48 – 2y – 2y = –2
48 – 4y = – 2

(Subtracting 48 from both sides of the equation gives)

48 – 4y – 48 = –2 –48 –4y = –50

(Dividing on both sides of the equation by – 4)

-4y/-4 = -50/-4
y = 50/4 = 25/2

(Putting this value of y into equation (2) and then solving for x gives)

x + 25/2 = 24

(Subtracting 25/2 from both sides of the equation gives)

x + 25/2 - 25/2 = 24 - 25/2
x = (48 - 25)/2 = 23/2

Hence the solution to the given system of equations is (x , y) = ( 23/2 , 25/( 2 ))

Note: Next we show what happens if we substitute the value of x into the same equation that we used to compute it (equation (2) in this example)

x + y = 24
24 – y + y = 24 ∵ (x = 24 – y)
24 = 24

This is the result that we are left with. There is nothing wrong with 24 being equal to 24, but then what should we do with it? Of course we have not been looking to prove this in the first place!!

Hence we conclude that there is no point in substituting the computed value into the same equation that was used for its computation. Always use the other equation!

As shown in the above example, we compute the variable value from one equation and substitute it into the other.

We are given that

y = 24 – 4x ------(1)
2x + y/2 = 12 ------(2)

Here we choose equation (1) to compute the value of x. Since equation (1) is already in its most simplified form:

(Putting this value of y into equation (2) and then solving for x gives)

2x + (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x)
2x + 24/2- 4x/2 = 12
2x + 12 – 2x = 12
12 = 12

You might feel that this is the same scenario as discussed above (that of 24 = 24). But wait! You are trying to jump at a conclusion a bit too early. In the previous scenario, the result 24 = 24 had resulted because we put the variable value into the same equation that we used for its computation. Here we have not done that.

The result 12 = 12 has got something to do with the nature of the system of equations that we are given. No matter what solving technique you might be using, a solution to a system of linear equations lies at a single point where their lines intersect. In this scenario, the two lines are basically the same (one line over the other. The following figure shows this scenario.

Such a system is called a dependent system of equations. And solution to such a system is the entire line (every point on the line is a point of intersection of the two lines)

Hence the solution to the given system of equations is the entire line: y = 24 – 4x

Another possible Scenario:

Similar to this example, there exists another scenario where substitution of one variable into the 2^nd equation leads to a result similar to one shown below:

23 = –46

Or

5 = 34

Such a scenario arises when there exists no solution to the given system of equations. I.e., when the two lines do not intersect at any point at all.

Hence in case of such a result, where your basic Math rules seem to fail, a simple conclusion is that no solution to the given system exists. Such a system of equations is called an Inconsistent system.