Solving Linear equations in One variable by Trial and Error Method

The linear equations in one variable can be solved by using trial and error method. In this method, different values of the variable are substituted in either one or both sides of the equation to check the property of the equality between them. If the trial is successful for a value, then the values of expressions in both sides of the equation are equal. Otherwise, it is considered as an error.

Due to testing the linear equation for different values, it is often called as Brute force method. Similarly, the value of the variable is actually guessed in this method. Therefore, it is also called as guessing method of solving linear equations in one variable.

Examples

The linear equations in one variable are mainly appeared in four mathematical forms. So, let’s learn all of them to understand how to solve linear equations in one variable mathematically by the trial and error method.

Addition form

$x+6 = 9$ is a linear equation in one variable.

$x$	$L.H.S$	$=$	$Value$	$Trial$
$0$	$0+6$	$=$	$6 \, (\ne 9)$	$Error$
$1$	$1+6$	$=$	$7 \, (\ne 9)$	$Error$
$2$	$2+6$	$=$	$8 \, (\ne 9)$	$Error$
$3$	$3+6$	$=$	$9$	$True$

Substitute different values in the place of $x$ in the left-hand side of equation and observe the value of expression. In this example, the trials from $x = 0$ to $x = 2$ are error.

For $x = 3$, the value of the left-hand side expression is equal to $9$ and it’s exactly equal to the right-hand side of the equation. So, the trials of guessing value of the variable can be stopped.

Therefore, the solution of the linear equation in one variable is $3$.

Subtraction form

$15-p = 20$ is a linear equation in one variable.

$p$	$L.H.S$	$=$	$Value$	$Trial$
$0$	$15-0$	$=$	$15 \, (\ne 20)$	$Error$
$-1$	$15-(-1)$	$=$	$16 \, (\ne 20)$	$Error$
$-2$	$15-(-2)$	$=$	$17 \, (\ne 20)$	$Error$
$-3$	$15-(-3)$	$=$	$18 \, (\ne 20)$	$Error$
$-4$	$15-(-4)$	$=$	$19 \, (\ne 20)$	$Error$
$-5$	$15-(-5)$	$=$	$20$	$True$

In this linear equation, the value of the right-hand side of the equation is $20$, which is more than $15$. If you take positive numbers, then the value of the left-hand side of the equation is less than $15$. So, it’s essential to take negative numbers in this case.

The trials from $x = 0$ to $x = -4$ are not correct but it is true for $x = -5$. Therefore, the root of this linear equation in one variable is $-5$.

Multiplication form

$9t = 27$ is a linear equation in one variable.

$t$	$L.H.S$	$=$	$Value$	$Trial$
$0$	$9 \times 0$	$=$	$0 \, (\ne 27)$	$Error$
$1$	$9 \times 1$	$=$	$9 \, (\ne 27)$	$Error$
$2$	$9 \times 2$	$=$	$14 \, (\ne 27)$	$Error$
$3$	$9 \times 3$	$=$	$27$	$True$

Do some trials by substituting different values in $t$ in the left-hand side of the equation and observe the value of the expression for each value.

From $t = 0$ to $t = 2$, the trials are error but it is true for $t = 3$.

Therefore, the solution $t$ equals to $3$ is known as the root of the linear equation in one variable.

Division form

$\dfrac{z}{3} = 2$ is a linear equation in one variable.

$t$	$L.H.S$	$=$	$Value$	$Trial$
$0$	$\dfrac{0}{3}$	$=$	$0 \, (\ne 2)$	$Error$
$1$	$\dfrac{1}{3}$	$=$	$0.3333 \, (\ne 2)$	$Error$
$2$	$\dfrac{2}{3}$	$=$	$0.6667 \, (\ne 2)$	$Error$
$3$	$\dfrac{3}{3}$	$=$	$1 \, (\ne 2)$	$Error$
$4$	$\dfrac{4}{3}$	$=$	$1.3333 \, (\ne 2)$	$Error$
$5$	$\dfrac{5}{3}$	$=$	$1.6667 \, (\ne 2)$	$Error$
$6$	$\dfrac{6}{3}$	$=$	$2$	$True$

Substitute different values in the place of $x$ in the left-hand side of the equation and get the value. If the value of quotient equals to the value in the right-hand side of the equation, then stop the process.

In this case, from $x = 0$ to $x = 5$, the value of the left-hand side of the equation is not equal to the value of the right-hand side of the linear equation. Therefore, all the trials are error. However, the value of the left-hand side of the equation is equal to $2$ if $x$ is equal to $6$.

Therefore, $x = 6$ is called as the root or solution of the linear equation in one variable.

Thus, the solutions of linear equations in one variable are calculated mathematically by trial and error method.

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