Algebra equations

To learn the math skill of “how to solve algebra equations” of the following various types, click on any of the links below:

  • Algebra equations of linear type in one variable
  • Algebra equations of linear type in two variables

Algebra equations of linear type in one variable:

Math problem 1:

Solve for x the linear equation:

5x = 30

Solution:

5x denotes product of 5 and x, i.e. 5 × x.

So, 5 × x = 30.

Now isolate x in the L.H.S. (Left Hand Side) by dividing by 5 on both the sides:

(5 × x)/5 = 30/5, so that the linear equation becomes

x = 6.

Hence the solution of the linear equation 5x = 30 is 5.

SHORT-CUT (for solving for x the linear equation 5x = 30) is

If 5x = 30, then one can transpose 5 to the R.H.S. directly by writing

x  = 30/5, which gives the solution for the linear equation as x = 6

Mixed Linear Equations in One Variable:

Math Problem No. 2:

Solve for x the following mixed linear equation:

6x + 7 = 31

Solution:

Step 1:

First transpose the constant term 7 from L.H.S. to the R.H.S by interchanging its sign thus:

6x = 31 – 7, so that the linear equation becomes

6x = 24.

Step 2:

Now transpose 6 in 6x which is same as 6 × x from L.H.S. to R.H.S.

The linear equation now converts into

x = 24/6, i.e. x = 4

thus the solution of the mixed linear equation 6x + 7 = 31 is 4.

Let us now, move onto learn the math skill of how to solve algebra equations of linear type in two variables.

1. Algebra equations of linear type in two variables

There are three methods to solve an algebra equation of linear kind in two variables. They are

 

  • The Method Of Substitution
  • The Method Of Elimination
  • The Method Of Cross Multiplication

Let us now learn the math skill of how to solve algebra equations of the linear type in two variables (linear equations in two variables are also called as Simultaneous Linear Equations)

The Method of Substitution:

Math Problem no. 3:

Solve for the two variables x and y the following simultaneous linear equations in x and y:

2x + 3y = 5 and 3x + 2y = 8

Solution:

The method of substitution (of solving simultaneous linear equations in two variables x and y) involves:

Expressing x in terms of y (i.e. one variable in terms of the other: x in terms of y or even y in terms of x) in one equation and substitution that algebraic expression in the other linear algebra equation.

It is thus:

From the 1st (linear algebra) equation 2x + 3y = 5, express x in terms of y thus

2x  = 5 – 3y, so that x = (5 – 3y)/2,

Now substitute this value (algebraic expression (5 – 3y)/2) of x in the second equation to ‘solve for y’ thus:

3 × (5 – 3y)/2 + 2y = 8

(15 – 9y)/2 + 2y = 8,

Now 2 is the L.C.M. (of the denominators in the L.H.S.) and after transposing it to R.H.S. as 8 × 2, we get (the following algebraic expression – a linear equation):

15 – 9y + 4y = 16,

-5y = 1, so y = -1/5

Now, substitute this value of y in either of the two above equations, say in the first, to solve for x as below:

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

2x = 5 + 3/5,

2x = 28/5, so that x = 14/5

Thus the solution of the system of simultaneous equations: 2x + 3y = 5 and 3x + 2y = 8, i.e. the values of x and y which satisfy the two algebra equations are
x= 14/5 and y = - 1/5.

The Method Of Elimination

(for solving algebra equations of linear type in two variables also called simultaneous linear equations)

 The algebra equations of linear type in two variables, also called as simultaneous linear equations, can also be solved by the method of elimination as discussed below:
2x + 3y = 5 and 3x + 2y = 8.

In this method, eliminate x to solve for y Or eliminate y to solve for x; i.e. eliminate one variable to solve for the other variable.

Learn this math skill as explained below:

2x + 3y = 5
3x + 2y = 8.

Let’s proceed with eliminating y to solve for x:

Multiply the 1st equation with the coefficient of y in the 2nd equation, i.e. 2 and multiply the 2nd equation with coefficient of y in the 1st, i.e. 3:

(2x + 3y = 5) × 2 and
(3x + 2y = 8) ×3

Rewrite the system of equations with the factors multiplied:

4x + 6y = 10 and 9x + 6y = 24.

Now subtract the L.H.S. and R.H.S. of second linear equation from those of the first linear equation as below:

(4x + 6y) – (9x + 6y) = 10 – 24

So that, 4x – 5x + 6y – 6y = -14

-x = -14, and finally x = 14.

Next to solve for y, either you can substitute x = 14 in any of the above two linear equations, or just to practice with the math skill of elimination method for solving simultaneous equations in two variables, we can proceed as below:

2x + 3y = 5
3x + 2y = 8.

Multiply the 1st linear equation with 3 – the x-coefficient in the 2nd equation and the 2nd linear equation with 2 – the x-coefficient in the 1st linear equation:

3 × (2x + 3y = 5) and 2 × (3x + 2y = 8).

Rewrite the system of equations with the factors multiplied as follows:

6x + 9y = 15 and
6x + 4y = 16.

Now, subtract the L.H.S. and the R.H.S. of the 2nd equation from the 1st equation as below:

(6x + 9y) – (6x +4y) = 15 – 16,

So, 6x – 6x + 9y – 4y = -1

5y = -1, finally y = -1/5

Related Links On Algebra Equations:

    • Algebra equations on absolute value
    • Algebra equations on quadratic equations
    • Algebra equations on bi-quadratics
    • Algebra equations on cubes
    • Algebra equations on exponents
    • Algebra equations on fractions
    • Algebra equations on logarithms
    • Algebra equations on radicals
    • Algebra equations on exponents

     



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