New Zealand
Level 6 - NCEA Level 1

# Check Solutions to Linear/Non-Linear Equations

Lesson

Whenever we believe we have found a solution to a mathematical problem, such as solving an equation, it is a good idea to check whether the proposed solution really is correct.

For example, suppose we are working with a formula $y=\frac{x-1}{2}+10$y=x12+10 and we wish to know what value of $x$x would give the result $y=25$y=25. Perhaps we have drawn a graph and it appears that $y=25$y=25 when $x$x is about $31$31. Or, perhaps we have worked through the algebra, as follows:

 $25$25 $=$= $\frac{x-1}{2}+10$x−12​+10 $15$15 $=$= $\frac{x-1}{2}$x−12​ $30$30 $=$= $x-1$x−1 $31$31 $=$= $x$x

We think we have the correct answer but to be sure we take the extra step of substituting $x=31$x=31 into the original equation to see whether it does indeed make $y=25$y=25. We put

 $y$y $=$= $\frac{31-1}{2}+10$31−12​+10 $=$= $\frac{30}{2}+10$302​+10 $=$= $15+10$15+10 $=$= $25$25

We see that the two sides of the equation are the same and, so,  the solution must be correct.

The same thing can be done with harder problems.

Consider the two equations $y=2x-1$y=2x1 and $y=3x-4$y=3x4. The equations represent two lines on a graph and because the lines have different gradients, they must cross somewhere.

It appears from the graph that the lines intersect where $x=3$x=3 and $y=5$y=5.  To check whether this is correct, we ask whether it is true that $2x-1=3x-4$2x1=3x4 when $x=3$x=3.

The left-hand side of the equation evaluates to $2\times3-1=5$2×31=5.

The right-hand side of the equation evaluates to $3\times3-4=5$3×34=5.

The two sides are the same. So, we have verified that for both lines, $y=5$y=5 when $x=3$x=3.

#### Example

Check whether $x=-4$x=4 is a solution to the quadratic equation  $x^2+\frac{5}{2}x-6=0$x2+52x6=0.

We evaluate the expression on the left of the equation with $x=-4$x=4 to see whether the result will match the right-hand side.

 $x^2+\frac{5}{2}x-6$x2+52​x−6 $=$= $(-4)^2+\frac{5}{2}\times(-4)-6$(−4)2+52​×(−4)−6 $=$= $16-10-6$16−10−6 $=$= $16-16$16−16 $=$= $0$0

The two sides are the same. So, $x=-4$x=4 is a solution.

#### Worked Examples

##### Question 1

Is $x=-15$x=15 a solution of $-\frac{2}{5}x=6$25x=6?

1. Yes

A

No

B

Yes

A

No

B

##### Question 2

Is $y=-2$y=2 a solution of $5y^2+4y=-12$5y2+4y=12?

1. Yes

A

No

B

Yes

A

No

B

##### Question 3

Is $x=\frac{1}{4}$x=14 a solution of $48x^2+16x-7=0$48x2+16x7=0?

1. Yes

A

No

B

Yes

A

No

B

### Outcomes

#### NA6-5

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

#### 91027

Apply algebraic procedures in solving problems