Question

if the length is increasing by 2m and breath is 3m then the area of rectangular floor of rafique is increasing by 75 sq m .but if the length is reduced by 2m and breath is increased by 3m the area is increased by 57 sq m .by forming simultaneously linear equation let us determine the length and breadth of the floor how to find the equation of this question​

Answer 1

Step-by-step explanation:

Let the length and breadth of the rectangle be X and Y. Length is increased by 2m and Breadth is increased by 3m then Area of the rectangle =>(X+2)(Y+3)=>XY+3X+2Y+6. So increase in Area =>XY+3X+2Y+6-(XY) =>2Y+3X+6= 75 => 2Y+3X=69...(1) But if Length is reduced by 2m and Breadth is increased by 3m there is decrease in area by 57 sq.m. =>(X-2)(Y+3) -XY= -57 =>XY+3X-2Y-6-(XY) = -57 =>3X-2Y-6= -57 =>3X-2Y= -57...(2) Add both equations (1) and (2) =>6X=12 =>X=2 and 3(2)-2Y= -57 =>2Y=63 =>Y= 31.5. Hence the length and breadth of the rectangle are 2m and 31.5m respectively. Hope it helps you.

Answer 2

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if the length is increasing by 2m and breath is 3m then the area of rectangular floor of rafique is increasing by 75 sq m .but if the length is reduced by 2m and breath is increased by 3m the area is increased by 15 sq m .by forming simultaneously linear equation let us determine the length and breadth of the floor how to find the equation of this question

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$\sf let \: the \: length \: be \:L m \: and \:breadth \: be \:b \: m$

$\sf so, \: the \:area \: is = Lb \:sq.m$

$\sf in \:case- I$

$\sf increased \: length= L+2$

$\sf increased \: breadth = b+3$

$\sf increased \:area = Lb+75$

$\sf\blue{To \: be \: noted:}$

$\sf {\boxed {\bf Length(L) {\bf × {\bf breadth (b) {\bf = {\bf area(Lb)}}}}}}$

$\sf (L+2)(b+3)=Lb+75$

$\sf Lb + 3L+ 2b+6=Lb+ 75$

$\sf Lb-Lb +3L+2b+6-75=0$

$\sf 3L+2b-69=0-------equation \: l$

$\sf in case- ll$

$\sf reduced \:length= L-2$

$\sf increased \: breadth = b+3$

$\sf increased \:area = Lb+15$

$\sf {\boxed {\bf Length(L) {\bf × {\bf breadth (b) {\bf = {\bf area(Lb)}}}}}}$

$\sf (L-2)(b+3)=Lb+15$

$\sf Lb+3L-2b-6=Lb+15$

$\sf Lb-Lb+3L-2b-6-15=0$

$\sf 3L-2b-21=0 -------- equation \:ll$

$\sf from \:both \: equations \: l \: and \:ll$

$\sf 3L+2b-69=0$

$\sf 3L-2b-21=0$

$\sf - \: + \: +$

_________$\sf (by \: subtracting)$

$\sf 0×3L+4b-48=0$

$\sf 4b=48$

b$\ = \dfrac{48}{4}$

$\sf b= 12$

$\sf putting \: the \: value \: of \:b \: in \:equation \: l$

$\sf 3L+2×12-69=0$

$\sf 3L+24-69=0$

$\sf 3L-45=0$

$\sf 3L=45$

L$\ = \dfrac{45}{3}$

$\sf L=15$

$\sf therefore, \: length \: is \:= \:15m \: and \:Breadth \: is \: = \: 12 \: m$