Question
(a) Find the integer values for x which satisfy the inequality \(–3 <2x –1 \leq 6\)
(b) Simplify \(\frac{x^2+3x-10}{x^2-25}\)
(c) (i) Show that \(\frac{5}{x-3}\frac{2}{x+1}=3\) can be simplified to \(3x^2-13x-8=0\)
(ii) Solve the equation \(3x^2-13x-8=0\)
Show all your working and give your answers correct to two decimal places.
Answer/Explanation
(a) Adding 1:
-3 + 1 < 2x – 1 + 1 \(\leq\) 6 + 1
-2 < 2x \(\leq\)7
Dividing by 2:
\(\frac{-2}{2} < \frac{2x}{2}\leq\frac{7}{2}\)
-1 < x \(\leq\)7/2
So, the integer values of x satisfying the inequality are:
x = 0, 1, 2, 3
Note: x cannot be 4 or greater than 4 because \(\frac{7}{2}\) is not an integer.
(b) \(x^2 + 3x – 10\) = (x + 5)(x – 2)
\(x^2 – 25\) = (x + 5)(x – 5)
Therefore,
\(\frac{(x^2 + 3x – 10)}{(x^2 – 25)} = \frac{(x + 5)(x – 2)}{(x + 5)(x – 5)}\)
On Solving,we get,
Simplified form of the expression is\(\frac {(x – 2)}{(x – 5)}
(c)(i) \(\frac{5}{x-3}\frac{2}{x+1}=3\)
= \(\frac{(5x + 5 + 2x – 6)}{(x-3)(x+1)}\)=3
= \(\frac{(7x – 1)}{(x-3)(x+1)}\)=3
Solving further we get,
7x – 1 = \(3(x^2 – 2x – 3)\)
7x – 1 = \(3x^2 – 6x – 9\)
which implies,
\(3x^2 – 13x – 8\) = 0
(ii) \(3x^2-13x-8=0\)
For solving, we use,
x = \(\frac{-b \pm \sqrt{b^2 – 4ac]}{2a}\)
where x is root of the equation
x = \(\frac{-(-13) \pm \sqrt{{-13}^2 – 4.3(-8)}}{2.3}\)
On solving we get,
x=4.88 and –0.55
Question
(a) Factorise completely.
\(3a^{2}b-ab^{2}\)
…………………………………………
(b) Solve the inequality.
3x+12< 5x-3
………………………………………….
(c) Simplify.
\((3x^{2}y^{4})^{3}\)
………………………………………….
(d) Solve.
\(\frac{2}{x}=\frac{6}{2-x}\)
x = …………………………………………
(e) Expand and simplify.
(x-2)(x+5)(2x-1)
………………………………………….
(f) Alan invests \($200\) at a rate of r% per year compound interest.
After 2 years the value of his investment is\( $206.46 .\)
(i) Show that \(r^{2}+200r-3232=0.\)
(ii) Solve the equation\( r^{2}+200r-323=0 \)to find the rate of interest.
Show all your working and give your answer correct to 2 decimal places.
r = …………………………………………
Answer/Explanation
(a) ab(3a – b) final answer
(b) x > 7.5 final answer
(c) \(27x^{6}y^{12}\)
(d)0.5 or \(\frac{1}{2}\)
(e) \(2x^{3}+ 5x^{2}– 23x + 10\) final answer
(f)(i) \(200\left ( 1+\frac{r}{100} \right )^{2}=206.46\)
\(1+\frac{2r}{100}+\frac{r^{2}}{100^{2}}\)
\(r^{2}+200r-323=0\)
(ii)\(\frac{-200+\sqrt{200^{2}-4(1)(323)}}{2\times 1}\)
1.60 final answer
Question
(a) Factorise completely.
\(3a^{2}b-ab^{2}\)
…………………………………………
(b) Solve the inequality.
3x+12< 5x-3
………………………………………….
(c) Simplify.
\((3x^{2}y^{4})^{3}\)
………………………………………….
(d) Solve.
\(\frac{2}{x}=\frac{6}{2-x}\)
x = …………………………………………
(e) Expand and simplify.
(x-2)(x+5)(2x-1)
………………………………………….
(f) Alan invests \($200\) at a rate of r% per year compound interest.
After 2 years the value of his investment is\( $206.46 .\)
(i) Show that \(r^{2}+200r-3232=0.\)
(ii) Solve the equation\( r^{2}+200r-323=0 \)to find the rate of interest.
Show all your working and give your answer correct to 2 decimal places.
r = …………………………………………
Answer/Explanation
(a) ab(3a – b) final answer
(b) x > 7.5 final answer
(c) \(27x^{6}y^{12}\)
(d)0.5 or \(\frac{1}{2}\)
(e) \(2x^{3}+ 5x^{2}– 23x + 10\) final answer
(f)(i) \(200\left ( 1+\frac{r}{100} \right )^{2}=206.46\)
\(1+\frac{2r}{100}+\frac{r^{2}}{100^{2}}\)
\(r^{2}+200r-323=0\)
(ii)\(\frac{-200+\sqrt{200^{2}-4(1)(323)}}{2\times 1}\)
1.60 final answer