Linear Equations with a Parameter

An equation can contain, besides the unknown $x$, another letter, such as $a$, that stands for a given number which can take different values. This letter is called a parameter. An equation with a parameter is really a whole family of equations, one for each value of the parameter, and solving it means finding its roots for every value of the parameter. Such equations are sometimes called parametric equations, but that name usually means the equations of a curve, such as ${x=\cos t}$, ${y=\sin t}$.

Every linear equation with a parameter can be brought to the form ${Ax=B}$, where $A$ and $B$ depend on the parameter. Then there are three cases:

Case	Roots
${A\ne 0}$	one root, ${x=\frac{B}{A}}$
${A=0}$ and ${B=0}$	every number is a root
${A=0}$ and ${B\ne 0}$	no root

So the key step is to find the values of the parameter that make the coefficient of $x$ zero and to check them separately. The lesson Solving Linear Equations shows the simplest cases.

Problem 1

Solve the equation for $x$:

A) ${x+a=7}$   B) ${2x+8a=4}$   C) ${x+a=2a-x}$
D) ${ax=5}$   E) ${a-x=x+b}$   F) ${ax=3a}$

Solution:

A) ${x=7-a}$. For every value of $a$ the equation has one root, ${7-a}$.

B) ${2x=4-8a}$, so ${x=2-4a}$.

C) ${{x+x}={2a-a}}$, so ${2x=a}$ and ${x=\frac{a}{2}}$.

D) If ${a\ne 0}$, divide by $a$: ${x=\frac{5}{a}}$. If ${a=0}$, the equation is ${0\cdot x=5}$, which has no root.

E) ${{a-b}={x+x}}$, so ${2x=a-b}$ and ${x=\frac{a-b}{2}}$.

F) If ${a\ne 0}$, divide by $a$: ${x=3}$. If ${a=0}$, the equation is ${0\cdot x=0}$, and every number is a root.

Problem 2

Solve the equation, where $a$ is a parameter:

A) ${(a+1)x=2a+3}$   B) ${2a+x=ax+4}$
C) ${a^2x-x=a}$   D) ${a^2x+x=a}$

Solution:

A) If ${a+1\ne 0}$, i.e. ${a\ne -1}$, then ${x=\frac{2a+3}{a+1}}$. If ${a=-1}$, the equation is ${0\cdot x=2\cdot(-1)+3}$, i.e. ${0\cdot x=1}$, which has no root.

B) ${{x-ax}={4-2a}}$, so ${(1-a)x=2(2-a)}$. If ${a\ne 1}$, then ${x=\frac{2(2-a)}{1-a}}$. If ${a=1}$, the equation is ${0\cdot x=2}$, which has no root.

C) ${(a^2-1)x=a}$, i.e. ${(a-1)(a+1)x=a}$. If ${a\ne 1}$ and ${a\ne -1}$, then ${x=\frac{a}{(a-1)(a+1)}}$. If ${a=1}$ or ${a=-1}$, the equation is ${0\cdot x=1}$ or ${0\cdot x=-1}$, which have no root.

D) ${(a^2+1)x=a}$. Since ${a^2\ge 0}$, the coefficient ${a^2+1}$ is at least 1 and never zero. So for every $a$ the root is ${x=\frac{a}{a^2+1}}$.

Problem 3

Solve the equation, where $a$ and $b$ are parameters:

A) ${ax+b=0}$   B) ${ax+2b=x}$
C) ${(b-1)y=1-a}$   D) ${(b^2+1)y=a+2}$

Solution:

A) ${ax=-b}$.

• If ${a\ne 0}$, the root is ${x=-\frac{b}{a}}$.
• If ${a=0}$ and ${b\ne 0}$, the equation is ${0\cdot x=-b}$ with ${-b\ne 0}$: no root.
• If ${a=0}$ and ${b=0}$, it is ${0\cdot x=0}$: every number is a root.

B) ${{ax-x}={-2b}}$, so ${(a-1)x=-2b}$.

• If ${a\ne 1}$, the root is ${x=-\frac{2b}{a-1}}$.
• If ${a=1}$ and ${b\ne 0}$, the equation is ${0\cdot x=-2b}$ with ${-2b\ne 0}$: no root.
• If ${a=1}$ and ${b=0}$, it is ${0\cdot x=0}$: every number is a root.

C)

• If ${b\ne 1}$, the root is ${y=\frac{1-a}{b-1}}$.
• If ${b=1}$ and ${a\ne 1}$, the equation is ${0\cdot y=1-a}$ with ${1-a\ne 0}$: no root.
• If ${b=1}$ and ${a=1}$, it is ${0\cdot y=0}$: every number is a root.

D) ${b^2+1\ge 1}$ for every $b$, so the coefficient of $y$ is never zero, and ${y=\frac{a+2}{b^2+1}}$ for all $a$ and $b$.

Problem 4

For which values of $x$ do the two expressions have equal values? Here $a$ is a parameter.

A) ${5x+a}$ and ${3ax+4}$
B) ${2x-2}$ and ${4x+5a}$

Solution: The expressions are equal for the roots of the equation you get by putting an equals sign between them.

A) ${{5x+a}={3ax+4}}$, so ${{5x-3ax}={4-a}}$ and ${(5-3a)x=4-a}$. If ${a\ne\frac{5}{3}}$, then ${x=\frac{4-a}{5-3a}}$. If ${a=\frac{5}{3}}$, the equation is ${0\cdot x=4-\frac{5}{3}}$, i.e. ${0\cdot x=\frac{7}{3}}$, which has no root: the expressions are never equal.

B) ${{2x-2}={4x+5a}}$, so ${{-2-5a}={4x-2x}}$ and ${2x=-2-5a}$. For every $a$, ${x=-\frac{2+5a}{2}}$.

Problem 5

Solve the equation, where $a$ is a parameter:

A) ${|ax+2|=4}$   B) ${|2x+1|=3a}$   C) ${|ax+2a|=3}$

Solution: An absolute value equation ${|X|=c}$ with ${c>0}$ splits into ${X=c}$ or ${X=-c}$. See absolute value equations.

A) ${ax+2=4}$ or ${ax+2=-4}$, i.e. ${ax=2}$ or ${ax=-6}$. If ${a\ne 0}$, the roots are ${x=\frac{2}{a}}$ and ${x=-\frac{6}{a}}$. If ${a=0}$, the equation is ${|2|=4}$, which is false: no root.

B) An absolute value is never negative.

• If ${a<0}$, the equation has no root.
• If ${a=0}$, it is ${|2x+1|=0}$, so ${2x+1=0}$ and ${x=-\frac{1}{2}}$.
• If ${a>0}$, then ${2x+1=3a}$ or ${2x+1=-3a}$, so ${x=\frac{3a-1}{2}}$ or ${x=-\frac{3a+1}{2}}$.

C) ${ax+2a=3}$ or ${ax+2a=-3}$, i.e. ${ax=3-2a}$ or ${ax=-3-2a}$. If ${a=0}$, the equation is ${|0|=3}$, which is false: no root. If ${a\ne 0}$, the roots are ${x=\frac{3-2a}{a}}$ and ${x=-\frac{3+2a}{a}}$.

Problem 6

Solve the equation ${2-x=2b-2ax}$, where $a$ and $b$ are parameters. For which integer values of $a$ is the root a natural number, if ${b=7}$?

Solution: Move the terms: ${{2ax-x}={2b-2}}$, i.e. ${(2a-1)x=2(b-1)}$.

• If ${a\ne\frac{1}{2}}$, the equation has one root, ${x=\frac{2(b-1)}{2a-1}}$.
• If ${a=\frac{1}{2}}$ and ${b=1}$, it is ${0\cdot x=0}$: every number is a root.
• If ${a=\frac{1}{2}}$ and ${b\ne 1}$, it is ${0\cdot x=2(b-1)}$ with ${2(b-1)\ne 0}$: no root.

Now let ${b=7}$. An integer $a$ is never ${\frac{1}{2}}$, so the root is

${x=\frac{2(7-1)}{2a-1}=\frac{12}{2a-1}}$

It is a natural number exactly when ${2a-1}$ is a positive divisor of 12. But ${2a-1}$ is odd, and the only odd positive divisors of 12 are 1 and 3:

${2a-1=1}$ gives ${a=1}$ (the root is 12), and ${2a-1=3}$ gives ${a=2}$ (the root is 4).

Answer: ${a=1}$ or ${a=2}$.

Problem 7

Solve the equation ${|ax-2-x|=4}$, where $a$ is a parameter. For which integer values of $a$ does the equation have a root that is a negative integer?

Solution: ${ax-2-x=4}$ or ${ax-2-x=-4}$, i.e.

${(a-1)x=6}$ or ${(a-1)x=-2}$

If ${a=1}$, these are ${0\cdot x=6}$ and ${0\cdot x=-2}$: no root. If ${a\ne 1}$, the roots are ${x=\frac{6}{a-1}}$ and ${x=-\frac{2}{a-1}}$. They have opposite signs, so at most one of them is negative.

• ${\frac{6}{a-1}}$ is a negative integer when ${a-1}$ is a negative divisor of 6: ${a-1=-1,-2,-3,-6}$, so ${a=0,-1,-2,-5}$.
• ${-\frac{2}{a-1}}$ is a negative integer when ${a-1}$ is a positive divisor of 2: ${a-1=1,2}$, so ${a=2,3}$.

Answer: ${a=-5,-2,-1,0,2,3}$.

Problem 8

Solve the equation:

A) ${3ax-a=1-x}$, where $a$ is a parameter;
B) ${2ax+b=2+x}$, where $a$ and $b$ are parameters.

Solution:

A) ${{3ax+x}={1+a}}$, so ${(3a+1)x=1+a}$. If ${a\ne -\frac{1}{3}}$, then ${x=\frac{1+a}{3a+1}}$. If ${a=-\frac{1}{3}}$, the equation is ${0\cdot x=1-\frac{1}{3}}$, i.e. ${0\cdot x=\frac{2}{3}}$, which has no root.

B) ${{2ax-x}={2-b}}$, so ${(2a-1)x=2-b}$. If ${a\ne\frac{1}{2}}$, then ${x=\frac{2-b}{2a-1}}$. If ${a=\frac{1}{2}}$, the equation is ${0\cdot x=2-b}$: if ${b=2}$, every number is a root, and if ${b\ne 2}$, there is no root.

Problem 9

The equation ${6(kx-6)+24=5kx}$ is given, where $k$ is an integer. For which values of $k$ does the equation

A) have the root ${-\frac{4}{3}}$;
B) have no roots;
C) have a root that is a natural number?

Solution: Simplify: ${{6kx-36+24}={5kx}}$, i.e. ${kx=12}$.

A) Put ${x=-\frac{4}{3}}$: ${-\frac{4}{3}k=12}$, so ${k=-9}$.

B) ${kx=12}$ has no root when ${k=0}$.

C) For ${k\ne 0}$ the root is ${x=\frac{12}{k}}$. It is a natural number when $k$ is a positive divisor of 12: ${k=1,2,3,4,6,12}$.

Problem 10

Solve the equation:

A) ${2ax+1=x+a}$, where $a$ is a parameter;
B) ${2ax+1=x+b}$, where $a$ and $b$ are parameters.

Solution:

A) ${{2ax-x}={a-1}}$, so ${(2a-1)x=a-1}$. If ${a\ne\frac{1}{2}}$, the only root is ${x=\frac{a-1}{2a-1}}$. If ${a=\frac{1}{2}}$, the equation is ${0\cdot x=\frac{1}{2}-1}$, i.e. ${0\cdot x=-\frac{1}{2}}$, which has no root.

B) ${{2ax-x}={b-1}}$, so ${(2a-1)x=b-1}$. If ${a\ne\frac{1}{2}}$, then ${x=\frac{b-1}{2a-1}}$. If ${a=\frac{1}{2}}$, the equation is ${0\cdot x=b-1}$: if ${b=1}$, every number is a root, and if ${b\ne 1}$, there is no root.

Problem 11

The equation ${3(ax-4)+4=2ax}$ is given, where the parameter $a$ is an integer. For which values of $a$ does the equation have a root that is

A) ${-\frac{2}{3}}$;
B) an integer;
C) a natural number?

Solution: Simplify: ${{3ax-12+4}={2ax}}$, i.e. ${ax=8}$.

A) Put ${x=-\frac{2}{3}}$: ${-\frac{2}{3}a=8}$, so ${a=-12}$.

B) If ${a=0}$, there is no root. If ${a\ne 0}$, the root is ${x=\frac{8}{a}}$, which is an integer when $a$ is a divisor of 8: ${a=\pm 1,\pm 2,\pm 4,\pm 8}$.

C) ${x=\frac{8}{a}}$ is a natural number when $a$ is a positive divisor of 8: ${a=1,2,4,8}$.

Problem 12

In the equation ${2-x=2b-2ax}$ from Problem 6, let ${b=7}$. For which integer values of $a$ is the root an integer?

Solution: As in Problem 6, the root is ${x=\frac{12}{2a-1}}$. It is an integer when ${2a-1}$ divides 12. Since ${2a-1}$ is odd, it can only be ${\pm 1}$ or ${\pm 3}$:

${2a-1}$	$a$	root ${\frac{12}{2a-1}}$
1	1	12
$-1$	0	$-12$
3	2	4
$-3$	$-1$	$-4$

Answer: ${a=-1,0,1,2}$. A natural root needs a positive divisor, which leaves only ${a=1}$ and ${a=2}$, as in Problem 6.

Problem 13

The function ${f(x)=(3a-1)x-2a+1}$ is given, where $a$ is a parameter. For which values of $a$ does the graph of the function

A) cross the $x$-axis;
B) not cross the $x$-axis?

Solution: The graph crosses the $x$-axis at the points where ${f(x)=0}$. So it crosses the axis exactly when the equation ${(3a-1)x-2a+1=0}$, i.e. ${(3a-1)x=2a-1}$, has a root.

A) If ${a\ne\frac{1}{3}}$, the root is ${x=\frac{2a-1}{3a-1}}$, and the graph crosses the $x$-axis at this point.

B) If ${a=\frac{1}{3}}$, the equation is ${0\cdot x=\frac{2}{3}-1}$, i.e. ${0\cdot x=-\frac{1}{3}}$, which has no root. Then ${f(x)=\frac{1}{3}}$ for every $x$: the graph is a horizontal line that does not cross the $x$-axis.

Problem 14

Solve the equation, where $a$ is a parameter:

A) ${|x-2|=a}$   B) ${|ax-1|=3}$   C) ${|ax-1|=a-2}$

Solution:

A)

• If ${a<0}$, there is no root.
• If ${a=0}$, then ${x-2=0}$ and ${x=2}$.
• If ${a>0}$, then ${x-2=a}$ or ${x-2=-a}$, so ${x=2+a}$ or ${x=2-a}$.

B) ${ax-1=3}$ or ${ax-1=-3}$, i.e. ${ax=4}$ or ${ax=-2}$. If ${a\ne 0}$, the roots are ${x=\frac{4}{a}}$ and ${x=-\frac{2}{a}}$. If ${a=0}$, the equation is ${|{-1}|=3}$, which is false: no root.

C)

• If ${a-2<0}$, i.e. ${a<2}$, there is no root.
• If ${a=2}$, the equation is ${|2x-1|=0}$, so ${x=\frac{1}{2}}$.
• If ${a>2}$, then ${ax-1=a-2}$ or ${ax-1=2-a}$, i.e. ${ax=a-1}$ or ${ax=3-a}$. Here ${a\ne 0}$, so the roots are ${x=\frac{a-1}{a}}$ and ${x=\frac{3-a}{a}}$.

Problem 15

For which values of the parameter are the two equations equivalent, that is, have the same roots?

A) ${\frac{x+m}{2}=1-m}$ and ${(-x-1)^2-1=x^2}$
B) ${\frac{x+m}{2}=1-m}$ and ${\frac{x-m}{3}=1-2m}$
C) ${|3-x|+x^2-5x+3=0}$ and ${ax+2a=1+x}$, where ${x>3}$

Solution:

A) Solve the second equation. Since ${(-x-1)^2=(x+1)^2}$,

${{x^2+2x+1-1}={x^2}}$, so ${2x=0}$ and ${x=0}$

The first equation gives ${x+m=2-2m}$, i.e. ${x=2-3m}$. The equations are equivalent when their roots are equal: ${2-3m=0}$, so ${m=\frac{2}{3}}$.

B) The first equation has the root ${x=2-3m}$. The second gives ${x-m=3-6m}$, i.e. ${x=3-5m}$. The roots are equal when ${2-3m=3-5m}$, so ${2m=1}$ and ${m=\frac{1}{2}}$.

C) For ${x>3}$ we have ${3-x<0}$, so ${|3-x|=x-3}$. The first equation becomes

${{x-3+x^2-5x+3}={0}}$, i.e. ${x^2-4x=0}$ and ${x(x-4)=0}$

so ${x=0}$ or ${x=4}$. Only ${x=4}$ satisfies ${x>3}$.

The second equation is ${(a-1)x=1-2a}$. If ${a=1}$, it is ${0\cdot x=-1}$: no root. If ${a\ne 1}$, its root is ${x=\frac{1-2a}{a-1}}$. The equations are equivalent when this root is 4:

${{\frac{1-2a}{a-1}}={4}}$, so ${1-2a=4a-4}$, ${5=6a}$ and ${a=\frac{5}{6}}$

More on equations: Solving Linear Equations · Equation Word Problems · Linear Equation Problems · Absolute Value Equations