Math Examples of System of Equations

Given the inequality equality is:
⇒ a + b ≤ 3;
On changing this equation by equality then the equation becomes:
So we can write the equation as:
⇒ a + b ≤ 3;
⇒ a + b = 3;
Now we find the value of one coordinate, so here in the given equation we can find the coordinates of b;
⇒ a + b = 3;
⇒ b = -a + 3;
          
For further solving we put different values of “a” so that we can find the “b” coordinates for plotting the graph.
If we put the value of a is -2 then we get:
Put a = -2;
⇒ b = - (2) + 3; 
On further solving we get:           
⇒ b = 1;
So when we put value of “a” is -2 then we get value of b is 1.
If we put the value of a is 0 then we get:
Put a = 0;
⇒ b = - (0) + 3;   
On further solving we get:         
⇒ b = 3;
So when we put value of “a” is 0 then we get value of b is 3.
If we put the value of a is 3 then we get:
Put a = 3;
⇒ b = - (3) + 3;             
On further solving we get:
⇒ b = 0;
So when we put value of “a” is 3 then we get value of b is 0.
If we put the value of a is 2 then we get:
Put a = 2;
⇒ b = - (2) + 3;             
On further solving we get:
⇒ b = 1;
So when we put value of “a” is 6 then we get value of b is -3.
So the coordinates are: (-2, 1), (0, 3), (3, 0) and (2, 1).
Now we see how to plot the graph with the help of these coordinates:

For plotting the graph we follow the steps given below:
Step1: First we draw two axes, one axis in vertical direction and other axis in horizontal direction.
Step2: We take all the coordinates of ‘a’ on horizontal axis.
Step3: And take coordinates of ‘b’ on vertical axis.
Step4: Then mark the coordinates in horizontal and vertical direction.
Step5: suppose we put the number -2 on horizontal and 1 on vertical, and join both the points and we apply same procedure for all remaining coordinates.
Step6: At last we obtained inequality graph.

 

Given the inequality equality is:

⇒ 2a + 3b ≤ 10;

First we change this inequality equation by equality so that we can find coordinates for graph.

On changing this equation by equality then the equation becomes:

So we can write the equation as:

⇒ 2a + 3b ≤ 10;

⇒ 2a + 3b = 10;

Now we find the value of one coordinate, here in the equation we can find coordinates of b;

⇒ 2a + 3b = 10;

⇒ 3b = -2a + 10;

On further solving we get the value of b is:

⇒ b = -2a + 10;

               3

For further solving we put different values of “a” so that we can find the “b” coordinates for plotting the graph.

If we put the value of a is -3 then we get:

Put a = -3;

⇒ b = -2a + 10;

               3

⇒ b = -2(-3) + 10;

                  3

⇒ b = 6 + 10;

              3

On further solving we get:

⇒ b = 16;

            3

⇒ b = 5.33;

So when we put value of “a” is -3 then we get value of b is 5.33.

If we put the value of a is -1 then we get:

Put a = -1;

⇒ b = -2a + 10;

               3

⇒ b = -2(-1) + 10;

                  3

⇒ b = 2 + 10;

              3

On further solving we get:

⇒ b = 12;

            3

⇒ b = 4;

So when we put value of “a” is -1 then we get value of b is 4.

If we put the value of a is 1 then we get:

Put a = 1;

⇒ b = -2a + 10;

               3

⇒ b = -2(1) + 10;

                  3

⇒ b = -2 + 10;

              3

On further solving we get:

⇒ b = 8;

           3

⇒ b = 2.66;

So when we put value of “a” is 1 then we get value of b is 2.66.

If we put the value of a is 2 then we get:

Put a = 2;

⇒ b = -2a + 10;

               3

⇒ b = -2(2) + 10;

                  3

⇒ b = -4 + 10;

              3

On further solving we get:

⇒ b = 6;

           3

⇒ b = 2;

So when we put value of “a” is 2 then we get value of b is 3.

If we put the value of a is 2 then we get:

Put a = 4;

⇒ b = -2a + 10;

               3

⇒ b = -2(4) + 10;

                  3

⇒ b = -8 + 10;

              3

On further solving we get:

⇒ b = 2;

           3

⇒ b = 0.66;

So when we put value of “a” is 2 then we get value of b is 3.

So the coordinates are: (-3, 5.33), (-1, 4), (1, 2.66), (2, 2) and (4, 0.66).

Now we see how to plot the graph with the help of these coordinates:

For plotting the graph we follow the steps given below:

Step1: First we draw two axes, one axis in vertical direction and other axis in horizontal direction.

Step2: We take all the coordinates of ‘a’ on horizontal axis.

Step3: And take coordinates of ‘b’ on vertical axis.

Step4: Then mark the coordinates in horizontal and vertical direction.

Step5: suppose we put the number -3 on horizontal and 5.33 on vertical, and join both the points and we apply same procedure for all remaining coordinates.

Step6: At last we obtained inequality graph.

Given the inequality equality is:
⇒2a + b ≤ 18;
First we change this inequality equation by equality so that we can find coordinates for finding the graph.
On changing this equation by equality then the equation becomes:
So we can write the equation as:
⇒2a + b ≤ 18;
⇒2a + b = 18;
Now we find the value of one coordinate, so here in the given equation we can find the coordinates of b;
⇒b = -2a + 18;
For further solving we put different values of ‘a’ so that we can find the ‘b’ coordinates for plotting the graph.
If we put the value of ‘a’ as ‘1’ then we get:
Put a = 1;
⇒b = -2(1) + 18;
⇒b = -2 + 18;
On further solving we get:
⇒b = 16;
So when we put value of “a” is 1 then we get value of b is 16.
If we put the value of a is 2 then we get:
Put a = 2;
⇒b = -2(2) + 18;
⇒b = -4 + 18;
On further solving we get:
⇒b = 14;
So when we put value of ‘a’ is 2 then we get value of b is 14.
If we put value of ‘a’ as 4 then we get:
Put a = 4;
⇒b = -2(4) + 18;
⇒b = -8 + 18;
On further solving we get:
⇒b = 10;
So when we put value of ‘a’ as 4 then we get value of ‘b’ as 10.
If we put the value of ‘a’ as 5 then we get:
Put a = 5;
⇒b = -2(5) + 18;
⇒b = -10 + 18;
On further solving we get:
⇒b = 8;
Now, put value of ‘a’ as ‘5’ then we get value of ‘b’ as 8.
So the coordinates are: (1, 16), (2, 14), (4, 10) and (5, 8).
Now we see how to plot the graph with help of these coordinates.
For plotting the graph we follow the steps given below:
Step1: First we draw two axes, one axis in vertical direction and other axis in horizontal direction.
Step2: We take all the coordinates of ‘a’ on horizontal axis.
Step3: And take coordinates of ‘b’ on vertical axis.
Step4: Then mark the coordinates in horizontal and vertical direction.
Step5: Suppose we put -4 on horizontal and 26 on vertical line, and join both the points. Apply the same procedure for all remaining coordinates.
Step6: At last we obtained inequality graph.

In the given pair of equations, we observe that a₁ = b₂ and a₂ = b₁, it is the special situation, in such cases; we will first add the two equations and get:
47x + 31y = 63 ----- (1)
+ 31x + 47y = 15 ------ (2)
78x + 78 y = 78,
Or 78 (x + y = 1),
Or x + y = 1 ---- (3)
Now on subtracting (1) from (2) we get:
    47x + 31y = 63, 
  -  31x + 47y = 15,
16x   - 16y = 48,
Now in this equation we take 16 common   and divide the whole equation by 16, we get:
16 * (x – y = 3),
Or x – y = 3 ------ (4)
 Now we add (3) and (4) equation and by this the variable ‘y’ will be eliminated and we get:
2x = 4,
Now we divide both sides of the equation by 2 and we get:
 Or x = 4/2,
Or x = 2,
Now to get the value of y, we put the value of x in x – y = 3 and we get
2 – y = 3,
Or – y = 3 – 2,
Or -y = 1,
Now we multiply both sides of the equation by -1 and we get:
Y = -1.
So the solution to the given Set of equations is (2, -1).

We have the pair of equation as: 11x + 15 y = -23 ---- (1)
 And 7x – 2y = 20 ------ (2)
 In the given pair of equations, we observe that neither a₁ = a₂ nor b₁ =b₂,
 In such a case we multiply (1) by 2 and (2) by 15 and get:
2* (11x + 15 y = -23)   ---- (3)
 And 15* (7x – 2y = 20) ------ (4)
Or we get:
22x + 30 y = -46 ----- (5)
And 105x – 30 y = 300 ----- (6)
Now we add (5) and (6) and get
22x + 30 y = -46,
 105x – 30 y = 300,            
127x = 254,
Or x = 254 / 127,
Or x = 2,
Now Here by eliminating y from both the equations we get the value of x = 2.
 We put the value of x = 2 in the given equation 7x – 2y = 20, we get:
7 * 2 – 2 y = 20,
14 – 2y = 20,
-2y = 20 – 14,
Or -2y = 6,
Now we divide both sides of the equation by -2 and we get :
 Or y = -6/2,
Or y = -3,
So the solution to the above equation is (2, -3).

We have the system of the equations as
2x + y = 7 ----- (1)
 4x – 3y = -1 ----- (2)
In the given pair of equations, we observe that the value of a₁ <> a₂ and b1 <> b₂, so we multiply (1) by 3 to get,
3* (2x + y = 7),
Or 6x + 3y = 21 ------ (3)
   4x – 3y = -1 ------- (4)
 Now we add (3) and (4) and we get:
  6x + 3y = 21,
   4x – 3y = -1,
        10x = 20,
Or x = 20/ 10,
Or x = 2,
Now we put this value of x = 2 in (1) and get
2 * 2 + y = 7,
Or 4 + y = 7,
Or y = 7 – 4 = 3.
So the solution to the above equation is (2, 3).

In the given pair of equations, we observe that neither a₁ = a₂ nor b₁ = b₂,
So we multiply (1) by 3 and (2) by 2 and get:
3* (3x + 2y = 11) ----- (3)
 2* (2x -3y = -10) ----- (4)
Or, we get:
9x + 6y = 33,
4x – 6y = -20,
13x = 13,
Now we bring 13 from left side to right side of the equation to get the value of ‘x’,
X = 13/13,
Or we get x = 1,
Here by eliminating y from both the equations we get the value of   x = 1.
 We put the value of x = 1 in the given equation 3x + 2y = 11, and get:
3* 1 + 2y = 11,
Or 2y = 11 – 9,
Or 2y = 8,
Or y = 4.
So the solution to the above equation is (1, 4).

In the above pair of equations we observe that the value of b₁ =- 5 and b₂ = 5,
So we will add the two equations and get:
   4x – 5y -20 = 0 ---- (1)
+ 3x + 5y -15 = 0 ----- (2)
 7x -35 = 0,
Or, 7x = 35,
Or, x = 35/7,
Or, x = 5,
Here by eliminating ‘y’ from both the equations we get the value of x = 5,
Now we put the value of ‘x’ in (1) we get,
4*5 - 5y = 20,
Or, 20 – 5y = 20,
Or, -5y = 0, i.e. y = 0.
So the solution to the above equation is (5, 0).

As we have two equations:
3x – 4y = 1 -----(1)
2x – 3y = 1---- (2)
Form (2) we get: 2x = 3y – 1
Now we find the value of ‘x’ as:
Or, x = (3y-1)/2,
This value of ‘x’ will be substituted in the equation 3x – 4y = 1,
So we get:
3 * (3y - 1)/2 – 4y = 1,
Multiplying the above given equation by 2, we get:
6 * (3y – 1) – 8y = 2,
18y – 6 – 8y = 2,
10y = 2 + 8,
Or, y = 10/ 10,
Or, we get y = 1.
Now putting the value of y = 1 in 3x – 4y = 1.
We get: 3x – 4 * 1 = 1,
Or, 3x – 4 = 1,
Or, 3x = 5,
Or, x = 5/3.
So, the solution is (5/3, 1 ).
Thus by the method of substitution, we first get the value of y = 1 and then the value of x = 5/3.

Here we have the Set of equations:
3x + y + 1 = 0 ----- (1)
2x – 3y + 8 = 0 ------ (2)
From (1) we get y = -3x – 1,
Now substituting the value of ‘y’ in (2) we get:
2x – 3(-3x -1) + 8 = 0,
Now this becomes a linear equation with one variable, so now we solve it and get the value of ‘x’ as follows:
Or, 2x + 9x + 3 + 8 = 0,
Or, 11x + 11 = 0,
Or, 11x = -11,
Or, x = -11/11 = -1.
Now putting the value of x = -1 in y = -3x – 1 we get the value of another variable ‘y’ as follows:
y = -3 * (-1) -1,
Or, y = 3 – 1 = 2,
So, x = -1 and y = 2,
Thus by the method of substitution, we first get the value of x = -1 and then the value of y = 2.

We have 2x + 3y = 8 ----- ( 1 )
                       x – 2y + 3 = 0 ------(2)
So from (2) we get: x = 2y – 3,
Now on putting the value of x in (1) we get the value of another variable ‘y’ as follows:
2 *(2y – 3) + 3y = 8,
Now this becomes a linear equation with one variable, so now we solve it and get the value of ‘y’ as follows,
Or, 4y – 6 + 3y = 8,
Or, 7y – 6 = 8,
Or, 7y = 8 + 6,
Or, 7y = 14,
Or, y = 14,
so x = 2y – 3  or
x = 2 * 14 – 3,
or x = 28 – 3 = 25. 
Thus by the method of substitution, we first get the value of y = 14 and then the value of x = 25.

From (1) we get: x = y – 1,
Now putting this value of ‘x’ in (2) we get
3 (y – 1) + 2y -12 = 0,
Now this becomes a linear equation with one variable, so now we solve it and get the value of ‘y’ as follows
Or, 3y – 3 + 2y – 12 = 0,
Or, y -15 = 0,
Or, y = 15,
Now we put the value of ‘y’ in (1) and we get the value of another variable ‘x’ as follows:
x – 15 + 1 = 0,
x – 14 = 0, or x = 14,
So we get x = 14, y = 15.

We have x – 2y = 8, so from this equation we derive the value of x and then substitute it in (1) equation:
    X = 8 + 2y,
 Now substituting this value of x in (1) we get,
2 * (8 + 2y) + 3y = 2,
Now this becomes a linear equation with one variable, so now we solve it and get the value of ‘y’ as follows
Or, 16 + 4y + 3y = 2,
Or, 16 + 7y = 2,
Or, 7y = 2 – 16,
Or, 7y = -14,
Or, y = -14/7,
Or, y = -2.
Now putting this value in x = 8 + 2y, we get the value of another variable ‘x’ as follows:
x = 8 + 2 * (-2),
= 8 – 4 = 4.

We have two equations, 
x + y = 5 ------ ( 1),
x – y = 7 --------( 2 ),
Now we solve (2) to get x = 7 + y,
Now substituting the value of ‘x’ in (1) we get:
7 + y + y = 5
Or 7 + 2y = 5
Now this becomes a linear equation with one variable, so now we solve it and get the value of ‘y’ as follows:
Or, 2y = 5 – 7,
Or, 2y = -2,
Or, y = -2/2,
Or, y = -1, putting it in x – y = 7, we get the value of another variable ‘x’ as follows:
x – (-1) = 7,
Or, x = 7 – 1 = 6,
So solution of the equation is (6, -1).

In this equation, we find that ‘x’ appears on the denominator, so we first multiply the equation by ‘x’ and we get:
x * 1/x + 5x = x * 3/x + 8x,
Or 1 + 5x = 3 + 8x,
Now we subtract ‘5x’ from both sides of the equation, we get:
1 + 5x -5x = 3 + 8x – 5x,
Or 1 = 3 + 3x,
Now we subtract 3 form both sides and get:
1 – 3 = 3 + 3x – 3,
Or -2 = 3x,
Or -2/3 = x.