**To solve linear equations with two variables**

Elimination method refers to multiplying the coefficients with a constant. In general, there are two broad classifications through which a pair of linear equations in two variables is solved. One classification is **Graphical solution**, in which the pair of simultaneous equations is solved by plotting the graphs of the given equations. In another classification, there are three different algebraic methods to solve simultaneous linear equations. These methods are namely:

– Substitution method

– Elimination method

– Cross-multiplication method.

In the upcoming discussion, the theory, examples and use of elimination method is discussed. The meaning of solving simultaneous linear equations is to reduce the pair of equations to a single linear equation in one variable and then solve it using normal methods.

In elimination method, the coefficients of the given equations are multiplied by a constant to make the coefficients of one variable same in both equations. Finally, any one variable remains at the end, which can then easily be solved.

Taking a general case of two** linear equations**:

\(ax + by\) = \(c\)………(1)

\(px + qy\) = \(r\)……….(2)

Multiplying eqn. (1) by p, we get,

\(apx + bpy\) = \(cp\) ………..(3)

Similarly, on multiplying eqn.(2) with \(a\), we get-

\(apx + aqy\) = \(ar\)………….(4)

As per the elimination method, the coefficient of x obtained in equation (3) and equation (4) is same.

In order to remove the variable \(x\) and get a linear equation in one variable, equation (4) is subtracted from equation (3). We get

Also, from equation (1) we get,

\(ax\) = \(c – by\)

\(\Rightarrow x \) = \(\frac{c – by}{a}\)

\(\Rightarrow x \) = \(\frac{c-b(\frac{cp – ar}{bp – aq})}{a}\)

\(\Rightarrow x \) = \(\frac{c(bp – aq) – b(cp – ar)}{a(bp – aq)}\)

\(\Rightarrow x \) = \(\frac{cbp – caq – bcp + bar}{a(bp – aq)}\)

\(\Rightarrow x \) = \(\frac{a(br – cq)}{a(bp – aq)}\)= \(\frac{br – cq}{bp – aq}\)

Now consider an example with application of elimination method:

Consider the equations:

\(2x + 7y\) = \(10\)…………….. (1)

\(3x + y\) = \(6\)………………… (2)

There can be two ways to solve this problem.

In the first way, multiply equation (1) by 3 and equation (2) by2, we get,

\(6x + 21y\) = \(30\)……………..(3)

\(6x + 2y\) = \(12\)……………….(4)

The coefficients the x in equation (3) and equation (4) are same i.e. 6.

Finally, subtract equation (4) from equation (3). We get-

\(6x + 21y – 6x – 2y\) = \(30 – 12\)

\(\Rightarrow 19y\) = \(18\)

\(\Rightarrow y\) = \(\frac{18}{19}\)

In order to get the value of x, the value of y is substituted in equation (2),

\(3x + \frac{18}{19}\) = \(6\)

\(\Rightarrow 3x\) = \(6 – \frac{18}{19}\)

\(\Rightarrow 3x\) = \(\frac{96}{19}\)

\(\Rightarrow x\) = \(\frac{96}{57}\) = \(\frac{32}{19}\)

In the second way, multiply equation (2) with 7,

\(21x + 7y\) = \(42\)………….(5)

And equation (1) is \(2x + 7y\) = \(10\)

Subtracting equation (1) from equation (5), we get

\(19x\) = \(32\)

\(\Rightarrow x\) = \(\frac{32}{19}\)

Substituting the value of \(x\) in Eqn. (1),

\(2(\frac{32}{19}) + 7y\) = \(10\)

\(\Rightarrow 7y\) = \(10 – \frac{64}{19}\)

\(\Rightarrow 7y\) = \(\frac{126}{19}\)

\(\Rightarrow y\) = \(\frac{18}{19}\)

Again, the same solution is obtained. So, choosing the equation is immaterial.

Ultimately, the solution obtained is same.

The elimination method is preferred over substitution method when it is easy to multiply the coefficient and add or subtract the equations to eliminate one of the variables. The final aim is to make it linear equation in one variable and then it can be solved easily.

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