Solve the following equation and verify your answer: x^2-(x+1)(x+… - Vidyadip

Question

Solve the following equation and verify your answer: $\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=6$

✓

Answer

$\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-(\text{x}^2+2\text{x}+\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-\text{x}^2-2\text{x}-\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{-3\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
By cross multiplication: $-3\text{x}-2=6(5\text{x}+1)$
$\Rightarrow-3\text{x}-2=30\text{x}+6$
$\Rightarrow-3\text{x}-30=6+2$ (By transposition)
$\Rightarrow-33\text{x}=8$
$\Rightarrow\text{x}=\frac{8}{-33}$
$=\frac{-8}{33}$
$\therefore\text{x}=\frac{-8}{33}$ Verification: $\text{L.H.S.}=\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}$
$=\frac{\Big(\frac{-8}{33}\Big)^2-\Big(\frac{-8}{33}+1\Big)\Big(\frac{-8}{33}+2\Big)}{\Big(\frac{-8}{33}\Big)+1}$
$=\frac{\frac{64}{1089}-\Big(\frac{25}{33}\times\frac{58}{33}\Big)}{\frac{-40+33}{33}}$
$=\frac{\frac{64}{1089}-\frac{1450}{1089}}{\frac{-7}{33}}=\frac{\frac{64-1450}{1089}}{\frac{-7}{33}}$
$=\frac{-1386}{1089}\times\frac{33}{-7}=6=\text{R.H.S.}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Linear Equations in One Variable→Linear Equations in One Variable — all question types→MATHS STD 8 question papers→Gujarat Board STD 8 question papers→Gujarat Board question paper generator→

Similar questions

The area of a trapezium with equal non-parallel sides is $168m^2$. If the lengths of the parallel sides are $36\ m$ and $20\ m$, find the length of the non-parallel sides.

It $x$ and $y$ vary inversely, fill in the following blanks:

$x$	$9$	$...$	$81$	$243$
$y$	$27$	$9$	$...$	$1$

It is known that for a given mass of gas, the volume $v$ varies inversely as the pressure $p$. Fill in the missing entries in the following table:

v (in $cm^3$)	$...$	$48$	$60$	$...$	$100$	$...$	$200$
p (in atmospheres)	$2$	$...$	$\frac{3}{2}$	$1$	$...$	$\frac{1}{2}$	$...$

The following chart gives the growth in height
(in terms of percentage) of boys and girls with their respective ages:

Age (in yr)	Boys	Girls
8	72%	77%
9	75%	81%
10	78%	84%
11	81%	88%
12	84%	91%
13	88%	95%
14	92%	98%
15	95%	99%
16	98%	99.5%
17	99%	100%
18	100%	100%

Draw the line graph of above data on the same sheet and answer the following question
(i) In which year both the boys and the girls achieve their maximum height?
(ii) Who grows faster at puberty (14-16 yr of age)?

A cylindrical container with diameter of base $56\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $32\ cm \times 22\ cm \times 14\ cm$. Find the rise in the level of the water when the solid is completely submerged.

$\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)-\frac{9}{2}$

Multiply the monomial by the binomial and find the value of $x = -1, y = 0.25$ and $ z = 0.05: -3 x\left(y^2+z^2\right) $

By selling a pair of earnings at a discount of $25\%$ on the marked price, a jeweller makes a profit of $16\%$. If the profit is $Rs. 48$, what is the cost price? What is the marked price and price at which the pair was eventually bought?

An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by $4\ cm$ and the base decreased by $2\ cm$, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Draw a histogram of the following data:

Class interval:	$10-15$	$15-20$	$20-25$	$25-30$	$30-35$	$35-40$
Frequency:	$30$	$98$	$80$	$58$	$29$	$50$