Equations

Lesson

Linear equations are those that can be written in the form $ax+b=0$`a``x`+`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 $ax+b$`a``x`+`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.

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

The relation $y=3x-2$`y`=3`x`−2 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.

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

$5-2x=2(2-x)$5−2`x`=2(2−`x`)

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$4−2`x` by expanding the bracket. Then, if we turn each side separately into a function, we have

$y=5-2x$`y`=5−2`x` and

$y=4-2x$`y`=4−2`x`

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

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

A statement like $5-2x=2(2+x)$5−2`x`=2(2+`x`), for example, can be transformed to read $1=4x$1=4`x` or $4x-1=0$4`x`−1=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=4+2x

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

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}$5−2`x`=10−4`x`2. 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.

Does the equation $5(1+x)=7+3x$5(1+`x`)=7+3`x` 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+3`x`, 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 |

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

One solution

AInfinitely many

BNo solutions

COne solution

AInfinitely many

BNo solutions

C

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

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

$3x=-7$3

`x`=−7A$-3x=13$−3

`x`=13B$-3x=-7$−3

`x`=−7C$3x=13$3

`x`=13D$3x=-7$3

`x`=−7A$-3x=13$−3

`x`=13B$-3x=-7$−3

`x`=−7C$3x=13$3

`x`=13DCan the equation be transformed into something of the form $x=k$

`x`=`k`, for some value of $k$`k`?Yes

ANo

BYes

ANo

BHence determine the number of solutions that satisfy the equation.

infinitely many

Ano solutions

Bone solution

Cinfinitely many

Ano solutions

Bone solution

C

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

Infinitely many

ANo solutions

BOne solution

CInfinitely many

ANo solutions

BOne solution

C

Form and use linear, quadratic, and simple trigonometric equations

Apply graphical methods in solving problems

Apply algebraic methods in solving problems