NZ Level 7 (NZC) Level 2 (NCEA)
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Identify Solutions to Linear Equations
Lesson

Linear equations are those that can be written in the form $ax+b=0$ax+b=0 after a suitable transformation. We would also call an equation linear if it could be written so that each side of the equation separately has the form $ax+b$ax+b, where $a$a or $b$b or both could possibly be zero.

The unknown quantity $x$x occurs with power $1$1 only, and  if the $0$0 in the transformed equation is replaced by a variable $y$y, then the resulting relation has a straight line graph.

Example 1

The equation $x+3=5-2x$x+3=52x is a linear equation because it can be manipulated into the form $ax+b=0$ax+b=0. If $2x$2x is added to both sides, we have $3x+3=5$3x+3=5. Then, when $5$5 is subtracted from both sides, we have $3x-2=0$3x2=0 which is an equation in the required form.

The relation $y=3x-2$y=3x2 has a linear graph with a gradient of $3$3 and $y$y-intercept $-2$2.

 

Linear equations can have no solutions, one solution or infinitely many solutions.

no solutions

Statements can be written that cannot possibly be true for any value of the variable. For example, 

$5-2x=2(2-x)$52x=2(2x)

Any attempt to collect the like terms leads to the absurd statement $5=4$5=4. We say that such an equation is inconsistent or contradictory.

Notice that the right-hand side of the equation can be transformed to $4-2x$42x by expanding the bracket. Then, if we turn each side separately into a function, we have

$y=5-2x$y=52x and

$y=4-2x$y=42x

The graphs of these relations never intersect because their gradients are the same. So, there cannot be a solution to $5-2x=4-2x$52x=42x.

one solution

When an equation can be expressed in the form $ax+b=0$ax+b=0 with $a$a and $b$b not both zero, there is exactly one solution.

A statement like $5-2x=2(2+x)$52x=2(2+x), for example, can be transformed to read $1=4x$1=4x or $4x-1=0$4x1=0. So, it must represent a linear equation. It is easy to check that it has the solution $x=\frac{1}{4}$x=14.

If we think of the two sides of the original statement as representing two functions 

y=5-2x
y=4+2x

we see that the graphs are intersecting straight lines. They cross where $x=\frac{1}{4}$x=14.

infinitely many solutions

Equations can be written in which the two sides are essentially the same. These statements are true whatever the value of the variable $x$x. Thus, such equations have infinitely many solutions.

For example, we might write $5-2x=\frac{10-4x}{2}$52x=104x2. The two sides are essentially the same. If we made separate relations representing the two sides and drew their graphs, we would find that the two lines coincide. Thus, there are infinitely many common points in the two graphs and, therefore, infinitely many solutions.

 

Example 2

Does the equation $5(1+x)=7+3x$5(1+x)=7+3x have no solutions, one solution or infinitely many solutions? If there is a unique solution, what is it?

If we think of the two sides as though they came from separate relations $y=5(1+x)$y=5(1+x) and $y=7+3x$y=7+3x, it is clear that the graph of the first has gradient $5$5 while the graph of the second has gradient $3$3. So, the two lines must intersect. The solution is the $x$x-value at the intersection.

It is clear that the solution is $x=1$x=1. We can also arrive at this conclusion algebraically.

$5(1+x)$5(1+x) $=$= $7+3x$7+3x
$5+5x$5+5x $=$= $7+3x$7+3x
$2x$2x $=$= $2$2
$x$x $=$= $1$1

Worked Examples

Question 1

How many solutions does the equation $8x-4x=4$8x4x=4 have?

  1. One solution

    A

    Infinitely many

    B

    No solutions

    C

    One solution

    A

    Infinitely many

    B

    No solutions

    C

Question 2

Consider the equation $5x-10=8x+3$5x10=8x+3.

  1. Which of the following steps can be reached by successive transformations of the equation?

    $3x=-7$3x=7

    A

    $-3x=13$3x=13

    B

    $-3x=-7$3x=7

    C

    $3x=13$3x=13

    D

    $3x=-7$3x=7

    A

    $-3x=13$3x=13

    B

    $-3x=-7$3x=7

    C

    $3x=13$3x=13

    D
  2. Can the equation be transformed into something of the form $x=k$x=k, for some value of $k$k?

    Yes

    A

    No

    B

    Yes

    A

    No

    B
  3. Hence determine the number of solutions that satisfy the equation.

    infinitely many

    A

    no solutions

    B

    one solution

    C

    infinitely many

    A

    no solutions

    B

    one solution

    C

Question 3

How many solutions does the equation $10\left(5+x\right)=10\left(x+5\right)$10(5+x)=10(x+5) have?

  1. Infinitely many

    A

    No solutions

    B

    One solution

    C

    Infinitely many

    A

    No solutions

    B

    One solution

    C

Outcomes

M7-7

Form and use linear, quadratic, and simple trigonometric equations

91257

Apply graphical methods in solving problems

91261

Apply algebraic methods in solving problems

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