Solve (a) Solve the inequality 2/5 * (x - 2) - 1/6 * (x + 5) <= 0 and show the solution on a number line (b) If a number x is chosen at random from the set of integers \{x / 5 <= x <= 25\} find the probability that the number chosen is prime or a multiple of 3

Question

$$\left. \begin{array} { l } { \frac { 3 } { ( x - 2 ) } - \frac { 1 } { 6 } ( x + 5 ) = 0 } \\ { \text { ( i ) T h a t a n i n s a n m a l a n s i n d m a l \right.$$

Solve for x

$x=-7$
$x=4$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $6\left(x-2\right)$, the least common multiple of $x-2,6$.

$$6\times 3-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $6$ and $3$ to get $18$.

$$18-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $-\frac{1}{6}$ and $6$ to get $-1$.

$$18-\left(x+5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-1$ by $x+5$.

$$18+\left(-x-5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-x-5$ by $x-2$ and combine like terms.

$$18-x^{2}-3x+10=0$$

Add $18$ and $10$ to get $28$.

$$28-x^{2}-3x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-3x+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+28$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-3$$ $$ab=-28=-28$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-28$.

$$1,-28$$ $$2,-14$$ $$4,-7$$

Calculate the sum for each pair.

$$1-28=-27$$ $$2-14=-12$$ $$4-7=-3$$

The solution is the pair that gives sum $-3$.

$$a=4$$ $$b=-7$$

Rewrite $-x^{2}-3x+28$ as $\left(-x^{2}+4x\right)+\left(-7x+28\right)$.

$$\left(-x^{2}+4x\right)+\left(-7x+28\right)$$

Factor out $x$ in the first and $7$ in the second group.

$$x\left(-x+4\right)+7\left(-x+4\right)$$

Factor out common term $-x+4$ by using distributive property.

$$\left(-x+4\right)\left(x+7\right)$$

To find equation solutions, solve $-x+4=0$ and $x+7=0$.

$$x=4$$ $$x=-7$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $6\left(x-2\right)$, the least common multiple of $x-2,6$.

$$6\times 3-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $6$ and $3$ to get $18$.

$$18-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $-\frac{1}{6}$ and $6$ to get $-1$.

$$18-\left(x+5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-1$ by $x+5$.

$$18+\left(-x-5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-x-5$ by $x-2$ and combine like terms.

$$18-x^{2}-3x+10=0$$

Add $18$ and $10$ to get $28$.

$$28-x^{2}-3x=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-x^{2}-3x+28=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-3$ for $b$, and $28$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 28}}{2\left(-1\right)}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 28}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 28}}{2\left(-1\right)}$$

Multiply $4$ times $28$.

$$x=\frac{-\left(-3\right)±\sqrt{9+112}}{2\left(-1\right)}$$

Add $9$ to $112$.

$$x=\frac{-\left(-3\right)±\sqrt{121}}{2\left(-1\right)}$$

Take the square root of $121$.

$$x=\frac{-\left(-3\right)±11}{2\left(-1\right)}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±11}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{3±11}{-2}$$

Now solve the equation $x=\frac{3±11}{-2}$ when $±$ is plus. Add $3$ to $11$.

$$x=\frac{14}{-2}$$

Divide $14$ by $-2$.

$$x=-7$$

Now solve the equation $x=\frac{3±11}{-2}$ when $±$ is minus. Subtract $11$ from $3$.

$$x=-\frac{8}{-2}$$

Divide $-8$ by $-2$.

$$x=4$$

The equation is now solved.

$$x=-7$$ $$x=4$$

Steps for Completing the Square

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $6\left(x-2\right)$, the least common multiple of $x-2,6$.

$$6\times 3-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $6$ and $3$ to get $18$.

$$18-\frac{1}{6}\left(x+5\right)\times 6\left(x-2\right)=0$$

Multiply $-\frac{1}{6}$ and $6$ to get $-1$.

$$18-\left(x+5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-1$ by $x+5$.

$$18+\left(-x-5\right)\left(x-2\right)=0$$

Use the distributive property to multiply $-x-5$ by $x-2$ and combine like terms.

$$18-x^{2}-3x+10=0$$

Add $18$ and $10$ to get $28$.

$$28-x^{2}-3x=0$$

Subtract $28$ from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-3x=-28$$

Divide both sides by $-1$.

$$\frac{-x^{2}-3x}{-1}=-\frac{28}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{3}{-1}\right)x=-\frac{28}{-1}$$

Divide $-3$ by $-1$.

$$x^{2}+3x=-\frac{28}{-1}$$

Divide $-28$ by $-1$.

$$x^{2}+3x=28$$

Divide $3$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{2}$. Then add the square of $\frac{3}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=28+\left(\frac{3}{2}\right)^{2}$$

Square $\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=28+\frac{9}{4}$$

Add $28$ to $\frac{9}{4}$.

$$x^{2}+3x+\frac{9}{4}=\frac{121}{4}$$

Factor $x^{2}+3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{11}{2}$$ $$x+\frac{3}{2}=-\frac{11}{2}$$

Subtract $\frac{3}{2}$ from both sides of the equation.

$$x=4$$ $$x=-7$$

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Example

Question

Solve for x

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square