Formulas for Linear Equation Problems

July 7, 2020

Formulas For Linear Equation Problem

A linear equation is also known as algebraic equation in which each term has an exponent of one. The graph representation of the equation shows a straight line. Standard form of linear equation is y = m x + b. Where, x is the variable and y, m, and b are the constants.

Formulas for Linear Equations & Definitions

A linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.
Standard form of linear equation is y = mx + b. Where, x is the variable and y, m, and b are the constants.

Formulas of Linear equations in one variable

A Linear Equation in one variable is defined as ax + b = 0
Where, a and b are constant, a ≠ 0, and x is an unknown variable
The solution of the equation ax + b = 0 is x = $– \frac{b}{a}$ . We can also say that $– \frac{b}{a}$ is the root of the linear equation ax + b = 0.

Formulas of Linear equations in two variable

A Linear Equation in two variables is defined as ax + by + c = 0
Where a, b, and c are constants and also, both a and b ≠ 0

Formulas for Linear equations in three variable

A Linear Equation in three variables is defined as ax + by + cz = d
Where a, b, c, and d are constants and also, a, b and c ≠ 0

Formulas and Methods to solve Linear equations

Substitution Method

Step 1: Solve one of the equations either for x or y.

Step 2: Substitute the solution from step 1 into the other equation.

Step 3: Now solve this equation for the second variable.

Elimination Method

Step 1: Multiply both the equations with such numbers to make the coefficients of one of the two unknowns numerically same.

Step 2: Subtract the second equation from the first equation.

Step 3: In either of the two equations, substitute the value of the unknown variable. So, by solving the equation, the value of the other unknown variable is obtained.

Cross-Multiplication Method

Suppose there are two equation,

$p_{1}x +q_{1}y = r_{1}$ ……..(1)

$p_{2}x +q_{2}y = r_{2}$ ……..(2)

Multiply Equation (1) with p₂

Multiply Equation (2) with p1

$p_{1}p_{2}x +q_{1}p_{2}y = r_{1}p_{2}$

$p_{1}p_{2}x +p_{1}q_{2}y = p_{1}r_{2}$

Subtracting,
$q_{1}p_{2}y – p_{1}q_{2}y = r_{1}p_{2} – p_{1}r_{2}$

or, y (q₁ p₂ – q₂p₁) = r₂p₁ – r₁p₂

Therefore, y = $\frac{r_{2}p_{1} – r_{1}p_{2} }{q_{1}p_{2} – q_{2}p_{1} }$

= $\frac{r_{1}p_{2} – r_{2}p_{1} }{q_{2}p_{1} – q_{1}p_{2} }$

where (p₁q₂ – p₂q₁) ≠ 0

Therefore, $\frac{y}{r_{1}p_{2} – r_{2}p_{1} } = \frac{1}{q_{2}p_{1} – q_{1}p_{2} }$ …(3)

Multiply Equation (1) with q₂

Multiply Equation (2) with q1

$p_{1}q_{2}x +q_{1}q_{2}y = r_{1}q_{2}$

$q_{1}p_{2}x +q_{1}q_{2}y = q_{1}r_{2}$

Subtracting,

$p_{1}q_{2}x – p_{2}q_{1}x = r_{1}q_{2} – q_{1}r_{2}$

or , x(p₁q₂ – p₂q₁) = (q₁r₂ – q₂r₁)

or, x = $\frac{q_{1}r_{2} – r_{1}q_{2}}{p_{1}q_{2} – p_{2}q_{1}}$

Therefore, $\frac{x}{q_{1}r_{2} – r_{1}q_{2}} = \frac{1}{p_{1}q_{2} – p_{2}q_{1}}$ … (4)

where (p₁q₂ – p₂q₁) ≠ 0

From equations (3) and (4), we get,

$\frac{x}{q_{1}r_{2} – r_{1}q_{2}} =\frac{y}{r_{1}p_{2} – p_{1}r_{2}} = \frac{1}{p_{1}q_{2} – p_{2}q_{1}}$

where (p₁q₂ – p₂q₁) ≠ 0

Note: Shortcut to solve this equation will be written as

$\frac{x}{q_{1}r_{2} – r_{1}q_{2}} =\frac{y}{r_{1}p_{2} – r_{2}p_{1} } = \frac{1}{q_{2}p_{1} – q_{1}p_{2} }$

which means,

x = $\frac{q_{1}r_{2} – r_{1}q_{2}}{q_{2}p_{1} – q_{1}p_{2} }$

y = $\frac{r_{1}p_{2} – r_{2}p_{1}}{q_{2}p_{1} – q_{1}p_{2}}$

Important Formulas of Linear Equations & key points and Formulas to Remember

Suppose, there are two linear equations: a₁x + b₁y = c₁ and a₂x + b₂y = c₂

Now,

(A) If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ , then there will be one solution, and the graphs will have intersecting lines.

(B) If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , then there will be numerous solutions, and the graphs will have coincident lines.

(C) If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}≠ \frac{c_{1}}{c_{2}}$ then there will be no solution, and the graphs will have parallel lines.