Elimination Method for Solving Equations

In math, Elimination method is used in algebra course. In algebra, we use elimination method for solving systems of equations. There are two easy methods for solving systems of equations. They are substitution method and elimination method. Elimination method is sometimes more convenient than the substitution method.

Elimination method is also known as Addition or subtraction method. In elimination method, one variable in the equation is eliminated by adding or subtracting the two equations.

In this method, we eliminate one of the unknowns.

Step 1 :- Multiply the given equations by suitable numbers so as to make the coefficients of one of the unknowns, numerically equal

Step 2 :- Add the new equations, if the numerically equal coefficients are opposite in sign, otherwise, subtract them.

Step 3 :- The resulting equation is linear in one unknown. Solve it to obtain the value of one of the unknowns.

Step 4 :- Substitute the value of this unknown in any of the given equations. Solve it to vet the value of the other unknown.

Along with Elimination method for solving equations, students can also learn solving simultaneous equations by substitution method.

Here is more examples on how to use elimination method for solving equations:

Example 1:-

Solve `5/x + 6y = 13, 3/x + 4y = 7`

Solution : The given equations are

$\frac{5}{x}$ + 6y = 13 …(i)

$\frac{3}{x}$ + 4y = 7 …(ii)

Multiplying (i) by 3 and (ii) by 5, we get

$\frac{15}{x}$ + 18y = 39 …(iii)

$\frac{15}{x}$ + 20y = 35 …(iv)

Subtracting (iv) from (iii), we get - 2y = 4 ⇒ y = - 2

Subtracting y = - 2 in (i), we get

$\frac{5}{x}$ + 6 × (- 2) = 13 ⇒ $\frac{5}{x}$ - 12 = 13 ⇒ $\frac{5}{x}$ = $\frac{25}{1}$ ⇒ 25x = 5 ⇒ x = $\frac{5}{25}$ = $\frac{1}{5}$

Hence, x = `1/5` and y = - 2 is the solution of the given equations

Example 2:

Solve $\frac{1}{7x}$ + $\frac{1}{6y}$ = 3, $\frac{1}{2x}$ - $\frac{1}{3y}$ = 5

Solution : The given equations are

$\frac{1}{7x}$ + $\frac{1}{6y}$ = 3 …(i)

$\frac{1}{2x}$ - $\frac{1}{3y}$ = 5 …(ii)

Putting $\frac{1}{x}$ = u and $\frac{1}{y}$ = v, these equations become

$\frac{1}{7u}$ + $\frac{1}{6v}$ = 3 ⇒ 6u + 7v = 126 …(iii)

$\frac{1}{2u}$ - $\frac{1}{3v}$ = 5 ⇒ 3u - 2v = 30 …(iv)

Multiplying (iv) by 2, we get 6u - 4v = 60 …(v)

Subtracting (v) from (iii), we get 11v = 66 ⇒ v = $\frac{66}{11}$ = 6

Subtracting v = 6 in (iii), we get 6u + 7 × 6 = 126 ⇒ 6u = 126 - 42 ⇒ 6u= 84 ⇒ u = $\frac{84}{6}$ = 14

Therefore u = 14 ⇒ $\frac{1}{x}$ = 14 ⇒ x = $\frac{1}{14}$

And, v = 6 ⇒ $\frac{1}{y}$ = 6 ⇒ y = $\frac{1}{6}$

Hence, x = `1/14` and y = `1/6` is the solution of the given equations

Students can also get help with Algbera homework problems involving Elimination method used for solving equations from the online tutors.