Theorem

When a pair of linear equations meets, is parallel, or coincides

Algebra • Equations • Simultaneous Equations

Formula

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} ⟺ exactly one solution

For

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

: the lines intersect in exactly one point when

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

; they are parallel with no solution when

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}

; and they coincide with infinitely many solutions when

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

How to use it

1For 'the lines intersect' or 'the system has a unique solution', impose only $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ — the constants are irrelevant.
2The constant ratio is what separates parallel from coincident, so check it only once the first two ratios are equal.
3Equivalently, eliminate one variable: a unique solution exists exactly when the surviving coefficient $a_{1} b_{2} - a_{2} b_{1}$ is non-zero.

For what value of

m

m x + 2 y - 5 = 0

and

3 x + y + 4 = 0

fail to intersect?

Step-by-step

1
$\frac{m}{3} = \frac{2}{1}$
Why: Parallel lines need the first two ratios to be equal.
2
$m = 6$
3
$\frac{6}{3} = \frac{2}{1} = 2 \neq = \frac{- 5}{4}$
Why: The constant ratio differs, so the lines are parallel and distinct rather than coincident.

Answer

m = 6

Watch out

Compare the ratios, not the coefficients:

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

is the condition, and the value that makes them equal is the one EXCLUDED. Rewrite both equations in the same

a x + b y + c = 0

form before comparing, or the signs will mislead.

Also known as

condition for unique solution, parallel lines condition, consistent system, simultaneous linear equations condition, when do two lines intersect