Home › Algebra and expressions

Solve linear equations for every value of a parameter

Check zero factors before canceling

For an equation $C (a) x = D (a)$ , dividing by $C (a)$ is valid only where it is nonzero. At every value where $C (a) = 0$ , inspect the original equation: $0 = 0$ gives all real x, while $0$ equal to a nonzero value gives no solution.

Problem

Solve each equation for every real value of parameter $a$ : (a) $(a - 7) (a - 3) x = (a + 1) (a - 7)$ ; (b) $a^{2} (x - 1) = 4 x + 3 a + 2$ .

Answer

(a)

a = 7

: any

x

;

a = 3

: none; otherwise

x = (a + 1) / (a - 3)

. (b)

a = - 2

: any

x

;

a = 2

: none; otherwise

x = (a + 1) / (a - 2)

The cases a=7 and a=-2 would be lost by canceling too early.

Step-by-step solution

1. Solve the first equation

At $a = 7$ , both sides are zero for every $x$ . If $a$ is not 7, cancel $a - 7$ to obtain $(a - 3) x = a + 1$ . At $a = 3$ this says $0 = 4$ , so there is no solution. Otherwise $x = (a + 1) / (a - 3)$ .

2. Rearrange the second equation

Move the x terms together: $(a^{2} - 4) x = a^{2} + 3 a + 2$ . Factor both sides to get $(a - 2) (a + 2) x = (a + 1) (a + 2)$ .

3. Check its special cases

At $a = - 2$ , both sides are zero, so every $x$ works. At $a = 2$ , the equation becomes $0 = 12$ , so none works. For all other $a$ , cancel $a + 2$ and divide by $a - 2$ to get $x = (a + 1) / (a - 2)$ .

Add to Chrome - solve your own

Screenshot any problem on your screen and get the steps. Chrome on a computer; free.