Letter Equation Solver

Introduction for letter equation solver:

In algebra equation solver, the solver solves the equations using the substitution and elimination method. Generally equations are written in the form of x and y variables. Instead of, we can use letters also example a, b and c etc. The equation is like 2a + 3b = c, where c is a constant value. The equation is solved step by step using the substitution and elimination method by the solver. First take two letter equation and solve to get a or b. The solver first find either a or b, and substitute first value of a or b in the next equation to get the solutions.


Letter equation solver's methods and steps:


Steps in letter equation for the solvers:

Substitution method for letter equation solvers:

First express the equation, b in terms of a in any one of the given equations.

Substitute this value of b in another equation.

Then the solvers solve the obtained equation in step 2 and find the value of a.

Solver, substitutes’ the value of a in equation step 1 taken and find the value of b.

Elimination method for letter equation solvers:

If necessary the solver multiply the given equation by suitable constants so as to make the coefficient of one of the variables, numerically equal.

Then the solver do next step numerically equal coefficient are opposite in sign, add the new equations otherwise subtract them.

Solver now solve the linear equation so obtained .this gives the value of first variable.

Then finally the solvers substitute the value of this variable in any of the given equations and find the value of the other variable.


Letter equation examples:


Example 1:

Solve the 4 equations: 3a + 2b = 7 and 2a + 2b = 6, x + y = 5 and 2x + y = 7

Solution:

3a + 2b = 7------------------ (1)

2a + 2b = 6----------------- (2)

Take first two equations

Multiply the equation (2) by -1

3a + 2b = 7 ---------- (1)

-2a – 2b = -6 ---------- (2)

a           = 1

Substitute x = 1 in (2)

2(1) + 2b = 6

2 + 2b = 6

2 + 2b – 2 = 6 -2 (now the solvers subtract 2 on both side)

2b = 4

`(2b)/(2) ` = `4/2` (now the solvers divide 2 on both side)

b = 2

Values are (a,b) is (1,2)

Example 2:

Solve the equation: a + b =5 and 2a + b =5.

Solution:

a + b =5 -------------- (1)

2a + b =5 ----------- (2)

Express in terms of x of first equation

a = b - 5

Here solver substitutes the value of x in (2)

2(b-5) +b =5

2b- 10 + b = 5

3b -10 = 5 (add 10 on both side)

3b + 10 -10 = 5 + 10

3b = 15

`(3b)/(3)` = ` (15)/(3)` (divide by 3 on both sides)

b = 5

Substitute b = 5 in (1)

a + 5 =5 (here the solver subtract 5 on both side)

a - 5 = 5-5

a = 0

Therefore (a,b) is (0,5)