Solve the equation $ x + 10 = 4 $

Solution

$ x + 10 = 4 $

$ x + 10 - 10 = 4-10 $ (subtract 10 from both sides)

$x = -6$

Check: with $x = – 6$ the left-hand side (LHS) of the original equation gives,

$$ - 6 + 10 = 4 $$

which equals the right-hand side (RHS) as required.

Solve the equation $ 3x + 10 = 4 - x $

Solution

$$ 3x + 10 = 4 - x $$ $ 3x + 10 - 10 = 4 - x - 10 $ (subtract $10$ from both sides)
$$ 3x = -6 - x $$ $ 3x + x = -6 - x + x $ (add $x$ to both sides)

$$ 4x = -6 $$

$ {\Large\frac{4x}{4}} = - {\Large\frac{6}{4}} $ (divide both sides by $4$, i.e. the coefficient of $x$)

$ x = -{\Large\frac{6}{4}}$

$x=-{\Large\frac{3}{2}}$ (express the RHS in its lowest terms)

Check: Substitute the solution into both the LHS and RHS.

LHS: $ 3 \times \Big(-{\Large\frac{3}{2}}\Big) + 10 = - {\Large\frac{9}{2}} + {\Large\frac{20}{2}} = {\Large\frac{11}{2}} $

RHS: $ 4 - \Big(- {\Large\frac{3}{2}}\Big) = {\Large\frac{8}{2}} + {\Large\frac{3}{2}} = {\Large\frac{11}{2}} $

As the LHS = RHS the solution has been verified.

Solve the equation $ 3(x + 10) = 4 (2-x) $

Solution

$ 3(x + 10) = 4(2 - x) $
$ 3x + 30 = 8 - 4x $ (expand the brackets)
$ 3x + 30 - 30 = 8 - 4x - 30 $ (subtract 30 from both sides)
$ 3x = -22 - 4x $
$ 3x + 4x = -22 - 4x + 4x $ (add 4$x$ to both sides)
$ 7x = -22 $

${\Large\frac{7x}{7}} = - {\Large\frac{22}{7}} $ (divide both sides by 7, i.e. the coefficient of $x$)

$x = -{\Large\frac{22}{7}}$

Solve the equation $ 2(4x + 1) = 7(x + 3) $

Solution

$ 2(4x + 1) = 7(x + 3) $
$ 8x + 2 = 7x + 21 $ (expand the brackets)
$ 8x - 7x = 21 - 2 $ (gather like terms and simplify)
$x = 19$

Solve the equation

$$ \frac{1}{x} + 1 = 5 $$

Solution

Note that this equation contains a fraction, i.e. ${\Large\frac{1}{x}}$, and so we cannot allow $x = 0$ as this would lead to division by $0$.

$ {\Large\frac{1}{x}} + 1 = 5 $

$ {\Large\frac{1}{x}} + 1 - 1 = 5 - 1 $ (subtract 1 from both sides)

$ {\Large\frac{1}{x}} = 4 $

$ {\Large\frac{x}{x}} = 4x $ (multiply both sides by $x$)

$ 1 = 4x $ (simplify the LHS)

$ 4x = 1 $ (transpose to obtain the variable on the LHS)

Solve the equation

$$ \frac{4x + 2}{3} + \frac{2x + 1}{2} = \frac{1}{6} $$

Solution

$ {\Large\frac{4x + 2}{3}} + {\Large\frac{2x + 1}{2}} = {\Large\frac{1}{6}} $

$ {\Large\frac{6(4x + 2)}{3}} + {\Large\frac{6(2x + 1)}{2}} = {\Large\frac{6}{6}} $ (multiply through by the LCM of $2$, $3$,and $6$, i.e. $6$)

$ 2(4x + 2) + 3(2x + 1) = 1 $

$ 8x + 4 + 6x + 3 = 1 $ (expand brackets)

$ 14x = - 6 $ (gather like terms and simplify)

$ x = - {\Large\frac{6}{14}} $ (divide both sides by $14$, i.e. the coefficient of $x$)

$x = -{\Large\frac{3}{7}}$

Solve the equation $ 6x + 3 = x + 3 $

Solution

$ 6x + 3 = x + 3 $
$ 6x = x $ (subtract $3$ from both sides)
$ 5x = 0 $ (subtract $x$ from both sides)
$x=0$

The equation $0x = 6$ has no solution as there is no value of $x$ which when multiplied by $0$ gives the answer $6$.

Solve the equation $4x = 3$ and comment on the answer with respect to the three constraints specified above.

Solution

If $ 4x = 3 $ then $ x = {\Large\frac{3}{4}} $

Considering the three conditions:

(i). $ {\Large\frac{3}{4}} > 0 $ and so this is a feasible answer for $x$.

(ii). Clearly $ {\Large\frac{3}{4}} \; $ is not an integer and so this is not a feasible answer for $x$.

(iii). $ x = {\Large\frac{3}{4}} $ lies outside the specified range and so this is not a feasible answer for $x$.

Solve the equation $(a - b)x = 1$ and state the conditions on $a$ and $b$ for which a solution exists.

Solution

if $(a - b)x = 1$ then $x = {\Large\frac{1}{a - b}}$

There is a problem with the solution if $a = b$ as this leads to division by $0$. Hence, the answer is valid provided $a - b \neq 0 ,\; i.e.\; a \neq b$.

So, not only are we required to solve an equation we should check if any conditions are placed on the solution.

Sketch the line $y = 2x - 1$

Solution

To obtain two points on the line we can choose any two $x$-values and calculate the corresponding $y$-values. However, it is more informative to identify the $x$ and $y$ intercepts.

For the $y$-intercept set $x = 0$ giving $y = -1$ and the point $(0, –1)$ on the line.

For the $x$-intercept set $y = 0$ and solve the equation $2x - 1 = 0$ giving $ x = {\Large\frac{1}{2}}$ and the point $(1/2,0)$ on the line.

Note that from inspection of the equation we can see that this line has slope $m = 2$ and $y$-intercept

Sketch the line $y = -3x + 6$

Solution

Identify the $x$ and $y$ intercepts.

For the $y$-intercept set $x = 0$ giving $y = 6$ and so $(0, 6)$ lies on the line.

For the $x$-intercept set $y = 0$ and solve the equation, $-3x + 6 = 0$ giving $x = 2$ and so $(2,0)$ lies on the line.

Find the gradient of the line that passes through the two points $(2,3)$ and $(3,5)$ and determine the equation of the line.

Solution

$$ \text{let}(x_1,y_1) = (2,3).$$

$$ \text{let}(x_2,y_2) = (3,5).$$

$$ \text{Gradient} \; m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{3 - 2}= \frac{2}{1} = 2. $$

Choose one of the points, say $(3, 5)$.

Determine the $y$-intercept. From the general form

$y = mx + c$

$5 = (2)(3) + c$

$c = -1$

The equation of the line is therefore $ y = 2x - 1$

To check the answer we must show that both points lie on the line.

When $x = 2, y = 2 \times 2 - 1 = 3$ as required

When $x = 3, y = 2 \times 3 - 1 = 5$ as required

Each record in a particular database has $50$ characters. The overhead for the header of the database is $20$ characters. Determine a formula for the total size of the file and calculate the database space required for $1000$ records.

Solution

Let $n$ denote the number of records.

Let $s$ denote the size of the file.

The number of characters in $n$ records is $50n$ and to this we add the overhead for the header of the database, i.e. $20$.

The size ($s$) of the file can then be calculated from the linear equation, $s = 50n + 20$.

If we create 1000 records then we need database space for $s = 50 × 1000 + 20 = 50020$ characters.

A radiator contains an $8$ litre mixture of water and antifreeze. The mixture contains $40\%$ antifreeze. It is required to have the radiator contain a mixture that is $60\%$ antifreeze

Some of the existing mixture is to be removed and replaced by pure anitfreeze ($100\%$ antifreeze).

Determine how much of the mixture should be drained and replaced by pure antifreeze

Solution

Diagrammatically,

Radiator contains $8$ litres of water/antifreeze mixture.

Mixture contains $40\%$ anitfreeze.

Let $x$ (in litres) represent the amount of the mixture to be removed (and added).

After removal the radiator contains $(8 - x)$ litres of mixture.

After removal the radiator contains $(8 - x)(0.4)$ litres of antifreeze.

Add $x$ litres of pure antifreeze

Requires 8 litres of water/antifreeze mixture to contain $60\%$ antifreeze, i.e. $(8)(0.6)$ litres of antifreeze.

The radiator will contain the same volume of mixture before and after the changes are made. Hence

$(8 - x)(0.40) + x(1.0) =8(0.6) $
$3.2 - 0.4x + x \;\;\;\;\;\;\;\;\;\;\;\;=4.8$
$0.6x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,=1.6$
$x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\,\;\; = {\Large\frac{1.6}{0.6}}$
$x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\,\;\; = {\Large\frac{8}{3}} \approx 2.67.$

Hence we need to drain $2.67$ litres of the mixture and replace this with pure antifreeze to ensure the mixture is $60\%$ antifreeze.

An example of a $2 \times 2$ system of simultaneous equations is given by

Solution

$ 2x + 3y = 7$

$ 3x + 4y = 9 $

It can easily be verified by substituting $x = -1$ and $y = 3$ in the left-hand side of each equation that these values solve both equations at the same time.

Consider the equations $ x+ 2y = -1 $   (1)
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\qquad\qquad\,\,4x - 3y = 18$   (2)

Rearrange equation (1) to give, $x = -1 -2y$

Substitute for $x$ into equation (2) to give

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4(-1-2y)-3y=18 $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -4-8y-3y=18 $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -11y=22 $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y = -2 $$

Now substitute $y = -2$ in the expression obtained earlier for $x$ i.e. $x = -1 - 2y$ .

We have $x = -1 -2(-2) = -1 + 4 = 3$.

Hence the solutions are $x = 3$, $y = - 2$.

This is the unique pair of values that satisfy both equations (1) and (2) at the same time.

Alternatively, we could have rearranged equation (1) for $y$ to obtain $y=-{\Large\frac{1}{2}}-{\Large\frac{1}{2}}x$ , which is substituted into equation (2) to find $x = 3$. Then substitute $x = 3$ in the expression for $y$ to obtain $y = -2$.

Once again consider the equations:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x + 2y = -1 $     (1)
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4x - 3y = 18 $     (2)

Step 1: Make the coefficient of $y$ the same in both equations.

Multiply equation (1) by 3: $ \;\;\; 3x + 6y = -3 $     (3)

Multiply equation (2) by 2: $ \;\;\; 8x - 6y = 36 $     (4)

Step 2: Eliminate $y$ and solve the resulting one variable equation for $x$.

Add (3) to (4): $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 11x \;\;\;\;\;\;\ =33 \;\; \implies x = 3$

Step 3: Substitute for $x$ in either equation (1) or (2) and solve for $y$.

Substitute $x = 3$ in equation (1) to obtain $y = -2$.

Hence solutions are $x = 3$, $y = - 2$.

Whichever method is used the solution should be checked by substituting the values for $x$ and $y$ into the original equations.

Consider the equations $2x + 3y = 6 $     (1)
$ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x + 3y = 1 $     (2)

Subtract equation (2) from (1): $ 0x + 0y = 5 $

There are no values of $x$ and $y$ that satisfy this equation at the same time and so the system has no solution.

In this case the system is said to be inconsistent.

Consider the equations $\;\;\;\;\; 2x+3y=6 $     (1)
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4x + 6y = 12 $    (2)

Multiply equation (1) by 2: $\; 4x + 6y = 12 $    (3)

Write down equation (2) $\;\;\;\; 4x + 6y = 12$    (2)

Subtract equation (2) from (3): $ 0x + 0y = 0 $

We can choose any values for $x$ and $y$ to satisfy this equation and so the system has infinitely many solutions.

The sum of two numbers is 64. The first number is three times the second. Find the two numbers.

Solution

Let $x$ be the first number.

Let $y$ be the second number.

From the statement: “The sum of two numbers is 64”:

$ x + y = 64 $    (1)

From the statement: “The first number is three times the second”:

$ x = 3y $    (2)

Rewriting the above gives the system

$ x + y = 64 $    (1)

$ x - 3y = 0 $    (2)

Method 1

From equation (2), $x = 3y$.

Substitute for $x$ in equation (1) to give $4y = 64$ so that $y = 16$.

From $x = 3y$ we then have that $x = 3 \times 16 = 48$.

Method 2

Eliminate $x$ using (1) – (2) giving $4y = 64$ so that $y = 16$.

Substitute for $y$ in equation (1) giving $x + 16 = 64$ and so $x = 48$.

Hence, the two numbers are 16 and 48 which we should check using the original equations

Next year a father will be twice the age of his son. Twenty years ago the father was the same age as the son is now. Determine how old the father and son are.

Solution

Let $x$ be the son’s age now.

Let $y$ be the father’s age now.

From the statement: “Next year a father will be twice the age of his son”:

$y + 1 = 2 (x + 1)$

From the statement: “Twenty years ago the father was the same age as the son is now”:

$y - 20 = x$

Rewriting the above gives the system

$ y - 2x = 1$     (1)

$ y - x = 20$     (2)

Applying Method 2 we eliminate $y$ using (1) – (2) giving $x = 19$.

Substitute for $x$ in equation (2) giving $y - 19 = 20$ and so $y = 39$.

Hence, the father is 39 years old and the son is 19 years old.

The same result will be obtained using Method 1 and this is left as an exercise.

Solve the equation $x^2 - 2x - 24 = 0$ using the quadratic formula.

Solution

Comparing our equation with the general form we see that $a = 1, b = -2$ and $c = -24$. The discriminant is

$$ b^2 - 4ac = (-2)^2 - 4 \times 1 \times(-24) = 100 > 0 $$

and so the equation has two distinct real roots. Substituting the values above into the formula, noting that we have already calculated the discriminant, yields

$ x = {\Large\frac{-(-2)\pm \sqrt{100}}{2 \times 1}} $

$ x = {\Large\frac{2 \pm \sqrt{100}}{2}} = {\Large\frac{2 \pm 10}{2}} = 1 \pm 5. $

$x = - 4$ or $ x = 6$

Check by substituting these values into the left-hand side of the original equation, i.e.

When $x = -4$ we have, $(-4)^2 - 2(-4) -24 = 16 + 8 - 24 = 0$ as required.

When $x = 6$ we have, $6^2 - 2(6) - 24 = 36 - 12 - 24 = 0$ as required

Solve the equation $2x^2 - 3x - 4 = 0$ using the quadratic formula.

Solution

Here $a = 2, b = -3$ and $c = -4$ so the discriminant is,

$b^2 - 4ac = (-3)^2 - 4 \times 2 \times(-4) = 41 > 0$

and so the equation has two distinct real roots.

Substituting the values above into the formula yields

$$ x = \frac{-(-3) \pm \sqrt{41}}{2 \times 2} $$ $$ x = \frac{3 \pm \sqrt{41}}{4} $$ $$ x = \frac{3 + \sqrt{41}}{4} \; \text{and} \; x = \frac{3 - \sqrt{41}}{4} $$

These are exact values of the solutions and a calculator can be used to obtain the corresponding decimal approximations,
$ x = 2.35$ and $ x = - 0.85$ to 2 decimal places.

Solve the equation $9x^2 - 12x + 4 = 0$ using the quadratic formula.

Solution

Here $a = 9, b = -12$ and $c = 4$ so the discriminant is,

$b^2 - 4ac = (-12)^2 - 4 \times 9 \times 4 = 144 - 144 = 0$

and so the equation has one (repeated) real root.

Substituting the values above into the formula yields

$ x = {\Large\frac{-(-12) \pm \sqrt{0}}{2 \times 9}} $

$ x = {\Large\frac{12}{18}} $

Solve the equation $x^2 - 2x + 10 = 0$ using the quadratic formula.

Solution

Here $a = 1, b = -2$ and $c = 10$ so the discriminant is,

$b^2 - 4ac = (-2)^2 - 4 \times 1 \times 10 = 4 - 40 = -36 < 0$

and so the equation has no real solution.

Solve the equation $x^2 + 9x + 18 = 0$ using factorisation.

Solution

We look for two numbers whose sum is the coefficient of $x$, i.e. $9$, and whose product is the constant term, i.e. $18$. Clearly the two numbers we require are $3$ and $6$.

Using this result we split the $9x$ term into $3x$ and $6x$. Now write,

$$ x^2 + 9x + 18 = 0 $$ $$ x^2 + 3x + 6x + 18 = 0 $$ $$ x(x + 3) + 6(x + 3) =0 $$ $$ (x + 3)(x + 6) = 0 $$ $$ x + 3 = 0 \;\text{or}\;x + 6 = 0 $$ $x = -3$ or $x = -6$

These answers must be checked by substituting them into the left-hand side of the original equation.

Note that in some cases, especially when the coefficient of the $x^2$ term is 1, it is possible to factorise quadratics simply by inspection. This is a skill which improves with practice. For example, here we can avoid some of the detail by, as before, spotting that the two numbers we need multiply to give $18$ and add to give $9$. By inspection, or using trial and error, we find that the only numbers that satisfy these constraints are $3$ and $6$. Returning to our equation we can directly write $(x + 3)(x + 6) = 0$ and solve as earlier.

Solve the equation $x^2 + 4x - 12 = 0$ using factorisation.

Solution

We need two numbers whose sum is $4$ and whose product is $-12$.

The two numbers we require are $-2$ and $6$.

Now split the $4x$ term into $-2x$ and $6x$ so that

$$ x^2 + 4x - 12 = 0 $$ $$ x^2 - 2x + 6x - 12 = 0 $$ $$ x(x - 2)+ 6 (x - 2) = 0 $$ $$ (x - 2)(x + 6) = 0 $$ $$ x - 2 = 0 \; \text{or} \; x + 6 = 0 $$ $x = 2$ or $x = -6$

Solve the equation $x^2 - 4x - 21 = 0$ using factorisation.

Solution

We need two numbers whose sum is $-4$ and whose product is $-21$.

The two numbers we require are $-7$ and $3$.

Now split the $-4x$ term into $-7x$ and $3x$ so that

$$ x^2 - 4x - 21 = 0 $$ $$ x^2 - 7x + 3x - 21 = 0 $$ $$ x(x - 7)+ 3(x - 7) = 0 $$ $$ (x - 7)(x + 3) = 0 $$ $$ x - 7 = 0 \; \text{or}\; x + 3 = 0 $$ $x = 7$ or $x = -3$

Solve the equation $x^2 - 13x + 42 = 0$ using factorisation.

Solution

We need two numbers whose sum is $-13$ and whose product is $42$.

The two numbers we require are $-6$ and $-7$.

Now split the $-13x$ term into $-6x$ and $-7x$ so that

$$ x^2 - 13x + 42 = 0 $$ $$ x^2 - 6x - 7x + 42 = 0 $$ $$ x(x - 6) - 7 (x - 6) = 0 $$ $$ (x - 6)(x - 7) = 0 $$ $$ x - 6 = 0 \;\text{or}\; x - 7 = 0 $$ $x = 6$ or $x = 7$.

Solve the equation $5x^2 + 27x + 10 = 0$ using factorisation.

Solution

We look for two numbers whose sum is 27 but note that their product is no longer simply given by the constant term as it was in previous examples. In this case the product of the two numbers is obtained by multiplying the coefficient of $x^2$ by the constant term. Hence, we require that the product is $5 \times 10 = 50$. Note that this is precisely the same approach as used previously except that then the coefficient of $x^2$ was 1.

Clearly the two numbers we require are $2$ and $25$.

Using this result we now split the $27x$ term into $2x$ and $25x$. Now write,

$ 5x^2 + 27x + 10 = 0 $

$ 5x^2 + 25x + 2x + 10 = 0 $

$ 5x(x + 5) + 2 (x + 5) = 0 $

$ (x + 5)(5x + 2) = 0 $

Solve the equation $6x^2 - 7x - 3 = 0$ using factorisation.

Solution

Here we look for two numbers whose sum is $-7$, and whose product is $-18$, i.e. $6 \times (-3)$, the coefficient of $x^2$ times the constant term. Clearly the two numbers we require are $-9$ and $2$.

Now split the $-7x$ term into $-9x$ and $2x$. and write,

$$ 6x^2 - 7x - 3 = 0 $$ $$ 6x^2 - 9x + 2x - 3 = 0 $$ $$ 3x(2x - 3) + 1(2x - 3) = 0 $$ $$ (2x - 3)(3x + 1) = 0 $$ $x = {\Large\frac{3}{2}}$ or $x = - {\Large\frac{1}{3}}$

If you encounter problems when factorising remember that the quadratic formula is always an option to consider.

Solve the equation $x^2 - 81 = 0$ using factorisation.

Solution

We note that the left-hand-side is the difference of two squares, so we have

$$ x^2 - 81 = 0 $$ $$ x^2 - 9^2 = 0 $$ $$ (x + 9)(x - 9) = 0 $$ $x = -9$ or $x = 9$

Solve the equation $x^2 - 3 = 0$ using factorisation.

Solution

We note that the left-hand-side is the difference of two squares as $ 3 = (\sqrt{3})^2 $, so we have

$ x^2 - 3 = 0 $

$ x^2 - (\sqrt{3})^2 = 0 $

$ (x + \sqrt{3})(x - \sqrt{3}) = 0 $

Solve the equation $x^4 - 3x^2 + 2 = 0$ using factorisation.

Solution

Although this equation is not a quadratic by introducing a new variable $y$ and setting $y = x^2$ we can transform our equation to a new equation with the structure of a quadratic.

We let $y = x^2$ so that the equation becomes $y^2 - 3y + 2 = 0$.

We can solve this equation for $y$, either by factorisation giving $(y - 2)(y - 1) = 0$, or by applying the quadratic formula to obtain the solution $y = 1$, $y = 2$

Note however, that this is the solution for $y$ and we actually require the solution for $x$.

The original equation in $x$ consisted of a fourth order (quartic) polynomial and so it must have four solutions.

To find $x$ we write $ y = x^2 \implies x = \pm \sqrt{y} $

When $y = 1$ we obtain $x = \pm 1$ and when $y = 2$ we have $ x = \pm \sqrt{2}$.

Hence, we have found the required four solutions for $x$: $1,-1, \sqrt{2},-\sqrt{2}$

Determine the solutions of: $ x + {\Large\frac{1}{x}} = 4$.

Solution

Note that here we cannot have $x = 0$ as this would lead to division by $0$.

Multiply through by $x$ to obtain

$$ x^2 + 1 = 4x $$ $$ x^2 - 4x + 1 = 0. $$

This equation does not factorise easily and so we use the quadratic formula.

$ x = {\Large\frac{4 \pm \sqrt{(-4)^2 - 4 (1)(1)}}{2(1)}} $

$ = {\Large\frac{4 \pm \sqrt{12}}{2}} $

$ = {\Large\frac{4 \pm 2 \sqrt{3}}{2}} $

Solve the equation, $$ \frac{1}{2(x + 3)} = \frac{x - 2}{(x + 4)(x - 3)} .$$

Solution

Note that to avoid division by zero we cannot have $x = -3$, $x = -4$ or $x = 3$.

Cross-multiply to obtain

$$ (x + 4)(x - 3) = 2(x + 3)(x - 2) $$ $$ x^2 + x - 12 = 2x^2 + 2x - 12 $$ $$ x^2 + x = 0 $$ $$ x(x + 1) = 0 $$ $x = 0$ or $x = -1$.

Alternatively we could multiply through by the LCM of the denominators, i.e. $2(x + 3)(x + 4)(x - 3)$ to obtain $(x + 4)(x - 3) = 2(x + 3)(x - 2)$ and proceed as above.

Solve the equation, $$ \frac{5x + 4}{x(x - 2)} = \frac{7}{(x - 2)} = 0 $$.

Solution

To avoid division by zero we cannot have either $x = 0$ or $x = 2$ as a solution.

The LCM of the denominators is $x(x - 2)$ and so we multiply through by $x(x - 2)$ giving

$$ 5x + 4 = 7x $$ $$ -2x + 4 = 0 $$ $$ x = 2 $$

However, as noted above, we cannot have $x = 2$ and so the equation has no solution.

Alternatively rewrite the equation as $$ \frac{5x + 4}{x(x - 2)} = \frac{7}{(x - 2)}. $$

Cross-multiplication leads to $(5x + 4)(x - 2) = 7x(x - 2)$ which simplifies to produce the same result as above.

Plot the quadratic equation $y = x^2 + 4x + 1$.

Solution

To find the roots of the quadratic, where it cuts the $x$-axis, we solve $x^2 + 4x + 1 = 0$ and obtain $x = -2 + \sqrt{3}$ - (here we used the quadratic formula).

The graph cuts the $y$-axis where $x = 0$, i.e. at $y = 1$.

More points on the graph can be obtained by forming a table: