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Equations

The solutions of an equation with one or more unknowns are those values that make the equation true, or "satisfy the equation," when they are substituted for the unknowns of the equation.  An equation may have no solution or one or more solutions.  If two or more equations are to be solved together, the solutions must satisfy all of the equations simultaneously.

There are two systems of equations.

 

Consistent System: A system, which could have two or more simultaneous linear equations is known as consistent if it has at least one solution.

 

Inconsistent System: A system of two simultaneous linear equations is said to be inconsistent if it has no solution at all.

 

Two equations of the form of a1x + b1y = c1 and a2x + b2y = c2 will have:

  • The equations have a unique solution if .
  • The equations have infinitely many solutions if .                  
  • The equations have no solutions if .

Linear Equations

Linear Equations with one unknown

By performing basic operations (such as adding, subtracting or multiplying both sides with the same number) we can get the unknown on one side and thereby solve it.

 

Illustration: Solve the equation (5x – 6)/3 = 4.

We can use the following steps:

Step 1: Multiply by 3 on both sides:   (5x – 6) / 3 = 4     5x – 6 = 12      

Step 2: Add 6 on both sides               5x = 12 + 6      =      18           

Step 3: Divide by 5 on both sides              x = 18/5

 

Linear Equations with two unknowns

First method. Solve the equations 3x + 2y = 11 and x – y = 2.

Solution. From the second equation, x = 2 + y. Substitute (2 + y) in first equation to get:

3(2 + y) + 2y = 11 6 + 3y + 2y = 11

6 + 5y = 11   5y = 5 y = 1.

If y = 1, then x = 2 + 1 = 3.

 

Second method.  Solve the equations 6x + 5y = 29 and 4x - 3y = - 6.

Solution. In this method we must equate the coefficients of either x or y. In this case, we see that the LCM of coefficients of y is 15. So we multiply first equation by 3 and second equation by 5 to get:

18x + 15y = 87

20x - 15y = -30.

Adding the two equations eliminates y. We get 38x = 57, or x = 57/38 = 3/2.

Finally, substituting 3/2 for x in one of the equations gives y = 4.

 





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