Question 514 Marks
The sum of the digits of the digits of two digit number is $5$. If the digits are reversed, the number is reduced by $27$. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place $y.$
Required no. $= 10y + x$
If the digit’s are reversed,
Reversed no. $= 10x + y$
According to the question,
$x + y = 5 \dots...(1)$
and,
$(10y + x) - (10x + y) = 27$
$9y - 9x = 27$
$y - x = 3\dots ...(2)$
Now,
Adding equation $(1)$ and $(2),$
$y - x = 3 \dots....(2)$
$+ y + x = 5 \dots....(1)$
$2y = 8$
$y = 4$
From $(1)$
$x + 4 = 5$
$x = 1$
Require no is
$10 (4) + 1 = 41$
View full question & answer→Question 524 Marks
Solve :$\frac{2 x y}{x+y}=\frac{3}{2}, \frac{x y}{2 x-y}=-\frac{3}{10}, x+y \neq 0$ and $2 x-y \neq 0$
Answer$ \frac{2 x y}{x+y}=\frac{3}{2}$
$ \Rightarrow \frac{x+y}{x y}=\frac{4}{3}$
$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{4}{3}\ldots . .(1)$
$ \frac{x y}{2 x-y}=-\frac{3}{10}$
$ \Rightarrow \frac{2 x-y}{x y}=-\frac{10}{3}$
$\Rightarrow-\frac{1}{x}+\frac{2}{y}=-\frac{10}{3}\ldots . . .(2)$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, equations $(1)$ and $(2)$ become
$u+v=\frac{4}{3}$ and $-u+2 v=-\frac{10}{3}$
$ \Rightarrow 3 u+3 v=4$ and $-3 u+6 v=-10$
Adding, We have
$9 \mathrm{v}=-6$
$ \Rightarrow \mathrm{v}=-\frac{6}{9}=-\frac{2}{3}$
$ \Rightarrow \frac{1}{y}=-\frac{2}{3}$
$ \Rightarrow \mathrm{y}=-\frac{3}{2}$
Substituting $y=-\frac{3}{2}$ in $(1)$, We have
$\frac{1}{x}-\frac{2}{3}=\frac{4}{3}$
$\Rightarrow \frac{1}{x}=\frac{6}{3}=2$
$\Rightarrow \mathrm{x}=\frac{1}{2}$
Hence, $x=\frac{1}{2} \quad$ and $y=-\frac{3}{2}$
View full question & answer→Question 534 Marks
Solve:$\frac{a}{x}-\frac{b}{y}=0, \frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
AnswerGiven equation are $\frac{a}{x}-\frac{b}{y}=0$ and $\frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the above system of equations become
$a u-b v+0=0$
$ a b^2 u+a^2 b v-\left(a^2+b^2\right)=0$
By cross$-$multiplication, we have
$\frac{u}{-b \times\left[-\left(a^2+b^2\right)\right]-a^2 b \times 0}=\frac{-v}{a \times\left[-\left(a^2+b^2\right)\right]-a b^2 \times 0}=\frac{1}{a \times a^2 b-a b^2 \times(-b)}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{-v}{-a\left(a^2+b^2\right)}=\frac{1}{a^3 b+a b^3}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{v}{a\left(a^2+b^2\right)}=\frac{1}{a b\left(a^2+b^2\right)}$
$ \Rightarrow u=\frac{b\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$ and $v=\frac{a\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$
$ \Rightarrow \mathrm{u}=\frac{1}{a}$ and $\mathrm{v}=\frac{1}{b}$
$ \Rightarrow \frac{1}{x}=\frac{1}{a}$ and $\frac{1}{y}=\frac{1}{b}$
$ \Rightarrow \mathrm{x}=\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$
View full question & answer→Question 544 Marks
Solve :$\frac{34}{3 x+4 y}+\frac{15}{3 x-2 y}=5, \frac{25}{3 x-2 y}-\frac{8.50}{3 x+4 y}=4.5$
AnswerLet $a=3 x+4 y$ and $b=3 x-2 y$
$\therefore \frac{34}{3 x+4 y}+\frac{15}{3 x-2 y}=5$
$\Rightarrow \frac{34}{a}+\frac{15}{b}=5\ldots . .(1)$
$\frac{25}{3 x-2 y}-\frac{8.50}{3 x+4 y}=4.5$
$\Rightarrow-\frac{8.50}{a}+\frac{25}{b}=4.5\ldots . . .(2)$
Multiply equation $(2)$ by $4$ , We get:
$-\frac{34}{a}+\frac{100}{b}=18\ldots . . .(3)$
Adding equation $(1)$ and $(3)$
$-\frac{34}{a}+\frac{100}{b}=18$
$+\frac{34}{a}+\frac{15}{b}=5$
$\frac{115}{b}=23$
$b=5$
$3 x-2 y=5\ldots . . .(4)$
Substituting $b=5$ in equation $(1)$, We get
$\frac{34}{a}+\frac{15}{b}=5$
$ \frac{34}{a}+\frac{15}{5}=5$
$ \frac{34}{a}=2$
$ 2 a=34$
$ a=17$
$3 x+4 y=17\ldots . . .(5)$
Subtracting equation $(5)$ from $(4),$ We get:
$3 x-2 y =5$
$-3 x+4 y =17$
$- -$
$-6 y =-12$
$y =2$
Substituting $y=2$ in equation $(4)$, We get
$3 x-2(2)=5$
$ 3 x=9$
$ x=3$
$\therefore$ Solution is $\mathrm{x}=3$ and $\mathrm{y}=2$.
View full question & answer→Question 554 Marks
Solve :$\frac{20}{x+y}+\frac{3}{x-y}=7 ,\frac{8}{x-y}-\frac{15}{x+y}=5$
Answer$\frac{20}{x+y}+\frac{3}{x-y}=7\ldots . .(1)$
$\frac{8}{x-y}-\frac{15}{x+y}=5\ldots . . .(2)$
Multiplying equation no $(1)$ by $8$ and $(2)$ by $3 .$
$\frac{160}{x+y}+\frac{24}{x-y}=56\ldots . . .(3)$
$-\frac{45}{x+y}+\frac{24}{x-y}=15\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$\frac{160}{x+y}+\frac{24}{x-y}=56$
$-\frac{45}{x+y}+\frac{24}{x-y}=15$
$\frac{205}{x+y}= 41$
$\frac{205}{x+y}=41$
$x+y=5\ldots . . . .(5)$
From $(1)$
$ \frac{20}{5}+\frac{3}{x-y}=7$
$ \frac{3}{x-y}=3$
$x-y=1\ldots . . .(6)$
Adding equation $(5)$ and $(6)$
$x+y=5$
$+ x-y=1$
$2 x=6$
$x=3$
From $(5)$
$3+y=5$
$ y=5-3$
$ y=2$
View full question & answer→Question 564 Marks
Solve :$\frac{3}{x}-\frac{2}{y}=0$ and $\frac{2}{x}+\frac{5}{y}=19$ Hence, find $'a\ '$ if $\mathrm{y}=\mathrm{ax}+ 3.$
Answer$\frac{3}{x}-\frac{2}{y}=0\ldots . . .(1)$
$\frac{2}{x}+\frac{5}{y}=19\ldots . . .(2)$
Multiplying equation no. $(1)$ by $5$ and $(2)$ by $2 .$
$\frac{15}{x}-\frac{10}{y}=0 ........(3)$
$\frac{4}{x}+\frac{10}{y}=38 .........(4)$
Adding $(3)$ and $(4),$
$ \frac{15}{x}-\frac{10}{y}=0$
$ +\frac{4}{x}+\frac{10}{y}=38$
$\frac{19}{x}=38$
$ x=\frac{1}{2}$
From $(1)$
$ 3\left(\frac{1}{2}\right)-\frac{2}{y}=0$
$ y=\frac{1}{3}$
$ \therefore y=a x+3$
$ \frac{1}{3}=a\left(\frac{1}{2}\right)+3$
$ \frac{a}{2}=-\frac{8}{3}$
$ a=-\frac{16}{3}$
View full question & answer→Question 574 Marks
Solve :$4 x+\frac{6}{y}=15$ and $3 x-\frac{4}{y}=7$. Hence, find $a$ if $y=a x-2$
Answer$4 x+\frac{6}{y}=15\ldots . .(1)$
$3 x-\frac{4}{y}=7\ldots . .(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $6$
$16 x+\frac{24}{y}=60\ldots . .(3)$
$18 x-\frac{24}{y}=42\ldots .(4)$
Adding $(3)$ and $(4),$ We get
$\begin{gathered}16 x+\frac{24}{y}=60,+ 18 x-\frac{24}{y}=42,34 x=102,x=3\end{gathered}$
Substituting $x=3$ in $ (1),$ We get
$4(3)+\frac{6}{y}=15$
$\Rightarrow \frac{6}{y}=15-12$
$\Rightarrow y=\frac{6}{3}=2$
Now,
$y=a x-2$
$ 2=a(3)-2$
$ 3 a=4$
$ a=\frac{4}{3}=1 \frac{1}{3}$
View full question & answer→Question 584 Marks
Solve :$5 x+\frac{8}{y}=19, 3 x-\frac{4}{y}=7$
Answer$5 x+\frac{8}{y}=19\ldots .(1)$
$3 x-\frac{4}{y}=7\ldots . .(2)$
Multiplying equation $(2)$ by $2$ , We get,
$6 x-\frac{8}{y}=14 \dots......(3)$
Adding $(1)$ and $(3)$, We get,
$\begin{gathered}5 x+\frac{8}{y}=19,+6 x-\frac{8}{y}=14,11 x=33,x=3\end{gathered}$
Substituting $x=3$ in equation $(1),$ We get
$5(3)+\frac{8}{y}=19$
$ 15+\frac{8}{y}=19$
$ \frac{8}{y}=19-15$
$ y=\frac{8}{4}$
$ y=2$
View full question & answer→Question 594 Marks
Solve the pairs of equations:$\frac{3}{x}+\frac{2}{y}=10, \frac{9}{x}-\frac{7}{y}=10.5$
Answer$\frac{3}{x}+\frac{2}{y}=10\ldots . .(1)$
$\frac{9}{x}-\frac{7}{y}=10.5\ldots . .(2)$
Multiplying equation $(1)$ by $3$ , We get
$\frac{9}{x}+\frac{6}{y}=30\ldots . . .(3)$
Subtracting (2) from (3), We get
$\frac{9}{x}+\frac{6}{y}=30$
$ \frac{9}{x}-\frac{7}{y}=10.5$
$ \frac{ - +-} {\frac{13}{y}=19.5}$
$ y=\frac{13}{19.5}=\frac{2}{3}$
From $(1),$
$\frac{3}{x}+\frac{2 \times 3}{2}=10$
$ \Rightarrow \frac{3}{x}+3=10$
$ \Rightarrow \frac{3}{x}=7$
$ \Rightarrow x=\frac{3}{7}$
View full question & answer→Question 604 Marks
Solve :$\frac{9}{x}-\frac{4}{y}=8 ,\frac{13}{x}+\frac{7}{y}=101$
Answer$\frac{9}{x}-\frac{4}{y}=8\ldots .(1)$
$\frac{13}{x}+\frac{7}{y}=101\ldots . .(2)$
Multiplying equation no. $(1)$ by $7$ and $(2)$ by $4 .$
$\frac{63}{x}-\frac{28}{y}=56\ldots . .(3)$
$\frac{52}{x}+\frac{28}{y}=404\ldots . . .(4)$
Adding equation $(3)$ and $(4)$
$\frac{63}{x}-\frac{28}{y}=56$
$+\frac{52}{x}+\frac{28}{y}=404$
$\frac{115}{x}=460$
$x=\frac{115}{460}=x=\frac{1}{4}$
From $(1)$
$9 \times\left(\frac{4}{1}\right)-\frac{4}{y}=8$
$ -\frac{4}{y}=-28$
$ y=\frac{1}{7}$
View full question & answer→Question 614 Marks
Solve :$\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}, \frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
AnswerGiven equations are $\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}$ and $\frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, the system of equations become
$\frac{3}{2} u+\frac{2}{3} v=-\frac{1}{3}$ and $\frac{3}{4} u+\frac{1}{2} v=-\frac{1}{8}$
$ \Rightarrow \frac{9 u+4 v}{6}=-\frac{1}{3}$ and $\frac{3 u+2 v}{4}=-\frac{1}{8}$
$\Rightarrow 27 u+12 v=-6$ and $24 u+16 v=-4$
$ \Rightarrow 27 u+12 v+6=0$ and $24 u+16 v+4=0$
$\Rightarrow \frac{u}{12 \times 4-16 \times 6}=\frac{-v}{27 \times 4-24 \times 6}=\frac{1}{27 \times 16-24 \times 12}$
$ \Rightarrow \frac{u}{48-96}=\frac{-v}{108-144}=\frac{1}{432-288}$
$ \Rightarrow \frac{u}{-48}=\frac{-v}{-36}=\frac{1}{144}$
$ \Rightarrow \frac{u}{-48}=\frac{v}{36}=\frac{1}{144}$
$ \Rightarrow u=\frac{-48}{144}=\frac{1}{3}$ and $v=\frac{36}{144}=\frac{1}{4}$
$ \Rightarrow \frac{1}{x}=-\frac{1}{3}$ and $\frac{1}{y}=\frac{1}{4}$
$ \Rightarrow \mathrm{x}=-3$ and $\mathrm{y}=4 .$
View full question & answer→Question 624 Marks
Solve, using cross$-$multiplication :$0.4 x-1.5 y=6.5 ,0.3 x+0.2 y=0.9$
AnswerGiven equation are $0.4 x-1.5 y=6.5$ and $0.3 x+0.2 y=0.9$
Comparing with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, We have
$a_1=0.4, b_1=-1.5, c_1=-6.5$ and $a_2=0.3, b_2=0.2, c_2=0.9$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x=\frac{(-1.5) \times(-0.9)-(0.2) \times(-6.5)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$ and $y=\frac{(-6.5) \times(0.3)-(-0.9) \times(0.4)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$
$ \Rightarrow x=\frac{1.35+1.3}{0.08+0.45}$ and $y=\frac{-1.95+0.36}{0.08+0.45}$
$ \Rightarrow x=\frac{2.65}{0.53}$ and $y=\frac{-1.59}{0.53}$
$ \Rightarrow x=5$ and $y=-3$
View full question & answer→Question 634 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients $:13x+ 11y = 70;11x + 13y = 74$
Answer$13x + 11y = 70 \dots...(1)$
$11x + 13y = 74 \dots...(2)$
Adding$ (1)$ and $(2)$
$13x + 11y = 70$
$+ 11x + 13y = 74$
$24x + 24y = 144$
Dividing by $24,$
$x + y = 6 \dots....(3)$
Subtracting $(2)$ from $(1)$
$13x + 11y = 70$
$- 11x + 13y = 74$
$- - - $
$2x - 2y = - 4$
Dividing by $2$
$x - y = - 2 \dots....(4)$
Adding equation $(3)$ and $(4)$
$x - y = - 2$
$+ x + y = 6$
$2x = 4$
$x = 2$
From $(3)$
$2 + y = 6$
$y = 4$
View full question & answer→Question 644 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$2 x-3 y-3=0, \frac{2 x}{3}+4 y+\frac{1}{2}=0$
Answer$2 x-3 y-3=0$
$\Rightarrow 2 x-3 y=3 \dots.....(1)$
$\frac{2 x}{3}+4 y+\frac{1}{2}=0$
Multiply by $6 ,$
$6 \times \frac{2 x}{3}+6 \times 4 y+\frac{1}{2} \times 6=0 \times 6$
$4 x+24 y=-3\ldots . . .(2)$
Multiplying equation no. $(1)$ by $8$
$16 x-24 y=24 \dots.....(3)$
Adding equation $(3)$ and $(2)$
$16 x-24 y =24$
$+4 x+24 y =-3$
$20 x =21$
$x =\frac{21}{20}$
From $(1)$
$\therefore 2\left[\frac{21}{20}\right]-3 y=3$
$ \therefore-3 y=3-\frac{21}{20}$
$ \therefore y=-\frac{3}{10}$
View full question & answer→Question 654 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{x-y}{6}=2(4-x),2 x+y=3(x-4)$
AnswerThe given pair of linear equations are
$ \frac{x-y}{6}=2(4-x)$
$ \Rightarrow x-y=12(4-x)$
$ \Rightarrow x-y=48-12 x$
$\Rightarrow 13 x-y=48\dots....(1) $[ On simplifying $]$
$ 2 x+y=3(x-4)$
$ \Rightarrow 2 x+y=3 x-12$
$\Rightarrow x-y=12\dots.....(2) [$ On simplifying$ ]$
Multiply equation $(2)$ by $13$ , We get,
$13 x-13 y=156\dots .....(3)$
Multiply equation $(2)$ by $13$ , We get,
$13 x-13 y=156$
Subtracting equation $(1)$ from $(3)$
$13 x-13 y=156$
$- 13 x-y=48$
$-+y$
$-12 y=108$
$y=-9$
Substituting $y=-9$ in equation $(1)$, we get
$13 x-(-9)=48$
$ \Rightarrow 13 x=39$
$ \Rightarrow x=3$
$\therefore$ Solution is $x=3$ and $y=-9$.
View full question & answer→Question 664 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{1}{5}(x-2)=\frac{1}{4}(1-y),26 x+3 y+4=0$
Answer$ \frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$ \Rightarrow 4(x-2)=5(1-y)$
$ \Rightarrow 4 x-8=5-5 y$
$\Rightarrow 4 x+5 y=13\dots.....(1)$
$26 x+3 y=-4\dots.....(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $5 .$
$12 x+15 y=39\dots.....(3)$
$130 x+15 y=-20\dots.....(4)$
Subtracting equation (4) from (3)
$12 x+15 y =39$
$-130 x+15 y =-20$
$- +$
$-118 x =59$
$x =-\frac{59}{118}$
$x =-\frac{1}{2}$
From $(1)$
$4\left(-\frac{1}{2}\right)+5 y=13$
$ 5 y=13+2$
$ y=3$
View full question & answer→Question 674 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{5 y}{2}-\frac{x}{3}=8,\frac{y}{2}+\frac{5 x}{3}=12$
AnswerThe given pair of linear equations are
$\frac{5 y}{2}-\frac{x}{3}=8$
$\Rightarrow-\frac{x}{3}+\frac{5 y}{2}=8\dots.....(1) [$ On Similifying $]$
$\frac{y}{2}+\frac{5 x}{3}=12$
$\Rightarrow \frac{5 x}{3}+\frac{y}{2}=12\dots.....(2) [$ On Similifying$ ]$
Multiply equation $(1)$ by $5 ,$ we get
$-\frac{5 x}{3}+\frac{25 y}{2}=40\dots......(3)$
Adding equation $(3)$ and $(2)$
$-\frac{5 x}{3}+\frac{25 y}{2}=40$
$ + \frac{5 x}{3}+\frac{y}{2}=12$
$\frac{26 y}{2}=52$
$ \Rightarrow 13 y=52$
$ \Rightarrow \mathrm{y}=4$
Substituting $y=4$ in equation $(1),$ We get
$-\frac{x}{3}+\frac{5(4)}{2}=8$
$ \Rightarrow-\frac{x}{3}=8-10$
$ \Rightarrow \mathrm{x}=6$
$\therefore$ Solution is $\mathrm{x}=6$ and $\mathrm{y}=4$.
View full question & answer→Question 684 Marks
Solve :$4 x+\frac{x-y}{8}=17,2 y+x-\frac{5 y+2}{3}=2$
Answer$4 x+\frac{x-y}{8}=17 ($Given$)$
$\Rightarrow 32 x+x-y=136$
$\Rightarrow 33 x-y=136 \dots......(1)$
$2 y+x-\frac{5 y+2}{3}=2 ($Given$)$
$\Rightarrow 6 y+3 x-5 y-2=6$
$\Rightarrow 3 x+y=8\dots.......(2)$
Adding equations $(1)$ and $(2),$ we get
$33 x-y =136$
$+ 3 x+y =8$
$36 x =144$
$x =4$
Substituting $x=4$ in equation $(2),$ We get
$3 \times 4+y=8$
$ \Rightarrow 12+y=8$
$ \Rightarrow y=8-12$
$ \Rightarrow y=-4$
$ \therefore$ Solution is $x=4$ and $y=-4$
View full question & answer→Question 694 Marks
Solve :$\frac{7+x}{5}-\frac{2 x-y}{4}=3 y-5, \frac{5 y-7}{2}+\frac{4 x-3}{6}=18-5 x$
Answer$\frac{7+x}{5}-\frac{2 x-y}{4}=3 y-5($Given$)$
$ \Rightarrow 4(7+x)-5(2 x-y)=20(3 y-5)$
$ \Rightarrow 28+4 x-10 x+5 y=60 y-100$
$\Rightarrow-6 x-55 y=-128\ldots . . .(1)$
$\frac{5 y-7}{2}+\frac{4 x-3}{6}=18-5 x ($Given$)$
$ \Rightarrow 3(5 y-7)+4 x-3=6(18-5 x)$
$ \Rightarrow 15 y-21+4 x-3=108-30 x$
$\Rightarrow 34 x+15 y=132 \dots.......(2)$
Multiplying equation $(1)$ by $34$ and equation $(2)$ by $6 $, We get
$-204 x-1870 y=-4352\dots.....(3)$
$204 x+90 y=792\dots......(4)$
Adding equation $(3)$ and $(4),$ We get
$-204 x-1870 y=-4352\dots.....(3)$
$+204 x+90 y =792$
$-1780 y =-3560$
$\Rightarrow y =2$
Substituting $y=2$ in equation $(1),$ We get
$-6 \mathrm{x}-55 \mathrm{x} 2=-128$
$ \Rightarrow-6 \mathrm{x}-110=-128$
$ \Rightarrow-6 \mathrm{x}=-18$
$ \Rightarrow \mathrm{x}=3$
$\therefore$ Solution is $x=3$ and $y=2$.
View full question & answer→Question 704 Marks
Solve :$11(x - 5) + 10(y - 2) + 54 = 0,7(2x - 1) + 9(3y - 1) = 25$
Answer$11( x - 5 ) + 10( y - 2 ) + 54 = 0($given$)$
$\Rightarrow 11x - 55 + 10y - 20 + 54 = 0$
$\Rightarrow 11x + 10y - 21 = 0$
$\Rightarrow 11x + 10y = 21\dots....(1)$
$7( 2x - 1 ) + 9(3y - 1) = 25($given$)$
$\Rightarrow 14x - 7 + 27y - 9 = 25$
$\Rightarrow 14x + 27y - 16 = 25$
$\Rightarrow 14x + 27y = 41\dots.....(2)$
Multiplying equation $(1)$ by $27$ and equation $(2)$ by $10,$ we get,
$297x + 270y = 567\dots....(3)$
$140x + 270y = 410\dots.....(4)$
Subtracting equation $(4)$ from equation $(3),$ we get
$157x = 157$
$\Rightarrow x = 1$ Substituting $x = 1$ in equation $(1),$ we get,
$11 x 1 + 10y = 21$
$\Rightarrow 10y = 10$
$\Rightarrow y = 1$
$\therefore $ Solution set is $x = 1$ and $y = 1.$
View full question & answer→Question 714 Marks
Solve for $\mathrm{x}$ and $\mathrm{y}$ :$4 x=17-\frac{x-y}{8}, 2 y+x=2+\frac{5 y+2}{3}$
AnswerThe given pair of linear equations are
$4 \mathrm{x}=17-\frac{x-y}{8}$
$\Rightarrow 33 x-y=136 \dots...(1)($ On Simplifying $)$
$2 y+x=2+\frac{5 y+2}{3}$
$\Rightarrow 3 x+y=8\dots...(2)($ On Simplifying$ )$
Multiply equation $(2)$ by $11$ , we get,
$33 x+11 y=88\dots...(3)$
Subtracting equation $(1)$ from $(3)$
$\begin{gathered}33 x+11 y=88,-33 x-y=136,-+-,12 y=-48,y=-4\end{gathered}$
Substituting $y=-4$ in equation $(1),$ we get :
$33 x-(-4)=136$
$ \Rightarrow 33 x=132$
$ \Rightarrow x=4$
$\therefore$ Solution is $\mathrm{x}=4$ and $\mathrm{y}=-4$.
View full question & answer→Question 724 Marks
Solve for $x$ and $y :\frac{y+7}{5}=\frac{2 y-x}{4}+3 x-5; \frac{7-5 x}{2}+\frac{3-4 y}{6}=5 y-18$
AnswerThe given pair of linear equation are
$\frac{y+7}{5}=\frac{2 y-x}{4}+3 x-5$
$\Rightarrow 55 x+6 y=128\dots....(1)($ On simplifying $)$
$\frac{7-5 x}{2}+\frac{3-4 y}{6}=5 y-18$
$\Rightarrow 15 x+34 y=132\dots....(2)($ On simplifying $)$
Multiply equation $(1)$ by $3$ and equation $(2)$ by $11 $, we get:
$165 x+18 y=384\dots....(3)$
$165 x+374 y=1452\dots.....(4)$
Subtracting $(4)$ from $(3)$
$\begin{gathered}165 x+18 y=384,- 165 x+374 y=1452,- - -,-356 y=-1068,y=3\end{gathered}$
Substituting $y=3$ in equation $(1),$ we get
$55 x+6(3)=128$
$ \Rightarrow 55 x=110$
$ \Rightarrow x=2$
$\therefore$ Solution is $\mathrm{x}=2$ and $\mathrm{y}=3$.
View full question & answer→Question 734 Marks
If $10y = 7x - 4$ and $12x + 18y = 1;$ find the values of $4x + 6y$ and $8y - x.$
Answer$ 10 y=7 x-4$
$ -7 x+10 y=-4$
$ -(7 x-10 y)=4$
$7 x-10 y=4\ldots(1)$
$12 x+18 y=1\ldots(2)$
Multiplying equation no. $(1)$ by $12$ and $(2)$ by $7 .$
$84 x-120 y=48\ldots . .(3)$
$84 x+126 y=7\ldots .(4)$
Substract equation $(3)$ and $(4)$
$84 x-120 y=48$
$ 84 x+126 y=7 $
$ - - -$
$-246 y=41$
$y=\frac{41}{-246}$
$y=-\frac{1}{6}$
From $(1)$
$ 7 x-10\left(-\frac{1}{6}\right)=4$
$ 7 x+\frac{5}{3}=4$
$ 7 x=4-\frac{5}{3}$
$ 7 x=\frac{7}{3}$
$ x=\frac{7}{3} \times 7$
$ x=\frac{1}{3}$
$ \therefore 4 x+6 y=4\left(\frac{1}{3}\right)+6\left(-\frac{1}{6}\right)=\frac{4}{3}-\frac{6}{6}=\frac{4}{3}-1=\frac{1}{3}$
$ \therefore 8 y-x=8\left(-\frac{1}{6}\right)-\frac{1}{3}=-\frac{4}{3}-\frac{1}{3}=-\frac{5}{3}$
View full question & answer→Question 744 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients $:41x + 53y = 135,53x + 41y = 147$
Answer$41x + 53y = 135 \dots...(1)$
$53x + 41y = 147 \dots...(2)$
Adding equation $(1)$ and $(2)$
$41x + 53y = 135$
$+ 53x + 41y = 147$
$94x + 94y = 282$
Dividing by $94,$
$x + y = 3 \dots....(3)$
Subtracting equation $(2)$ from $(1)$
$41x + 53y = 135$
$- 53x + 41y = 147 $
$- - - $
$- 12x + 12y = - 12$
Dividing by $12,$
$- x + y = -1\dots....(4)$
Adding $(3)$ and $(4)$
$x + y = 3$
$+ - x + y = -1$
$2y = 2$
$y = 1$
From $(3)$
$x + y = 3$
$x + 1 = 3$
$x = 2$
View full question & answer→Question 754 Marks
Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution $:2( x - 3 ) + 3( y - 5 ) = 0,5( x - 1 ) + 4( y - 4 ) = 0$
AnswerGiven equations are
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3),$ we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.
View full question & answer→Question 764 Marks
Solve the following pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$:1.5x + 0.1y = 6.2,3x - 0.4y = 11.2$
AnswerThe given pair of linear equations are
$1.5 x+0.1 y=6.2\ldots(1)$
$3 x-0.4 y=11.2\ldots(2)$
equation $(1)$ dividing from $10$
$\frac{1.5 x}{10}+\frac{0.1 y}{10}=\frac{6.2}{10}$
$ \Rightarrow 15 x+y=62$
$y=62-15 x\ldots..(3)$
Putting the value of $y$ from in equation $(2)$
$3 x-0.4(62-15 x)=11.2$
$ \Rightarrow \frac{30 x-4(62-15 x)}{10}=\frac{112}{10}$
$ \Rightarrow 30 x-248+60 x=112$
$ \Rightarrow 90 x-248=112$
$ \Rightarrow 90 x=360$
$\Rightarrow x=4 \dots....(4)$
Substitute the value of $x$ from equation $(4)$ in equation $(3)$
$y=62-15 x$
$ =62-15(4)$
$ =62-60$
$ \Rightarrow y=2$
$\therefore$ Solution is $\mathrm{x}=4$ and $\mathrm{y}=2$.
View full question & answer→Question 774 Marks
Solve the following pair of linear $($simultaneous$)$ equation by the method of elimination by substitution:$0.2x + 0.1y = 25;2(x - 2) - 1.6y = 116$
AnswerThe given pair of linear equations are
$0.2 x+0.1 y=25\ldots .(1)$
$2(x-2)-1.6 y=116 \dots..(2)$
Consider equation $(1)$
$0.2 x+0.1 y=25$
$ \Rightarrow 0.2 x=25-0.1 y$
$\Rightarrow x=\frac{25-0.1 y}{0.2}\ldots .(3)$
Substitute the value of $x$ from equation $(3)$ in equation $(2).$
$2(x-2)-1.6 y=116$
$ \Rightarrow 2\left[\frac{25-0.1 y}{0.2}-2\right]-1.6 y=116$
$ \Rightarrow 2 \times \frac{25-0.1}{0.2}-2 \times 2-1.6 y=116$
$ \Rightarrow 0.2 \times \frac{50-0.2}{0.2}-4-1.6 y=116$
$ \Rightarrow 50-0.2 y-0.8-0.32 y=23.2$
$ \Rightarrow 26=0.52 y$
$\Rightarrow y=50 \dots....(4)$
Substitute the value of $y$ from equation $(4)$ in equation $(3).$
$x=\frac{25-0.1 y}{0.2}$
$ \Rightarrow x=\frac{25-0.1(50)}{0.2}$
$ \Rightarrow x=\frac{25-5}{0.2}$
$ \Rightarrow x=100$
$\therefore$ Solution is $x=100$ and $y=50$.
View full question & answer→Question 784 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution :$8x + 5y = 9,3x + 2y = 4$
Answer$ 8 x+5 y=9$
$3 x+2 y=4 \ldots(1)$
$8 x+5 y=9$
$\therefore 5 y=9-8 x$
$\therefore y=\frac{9-8 x}{5} \ldots .(2) $
Putting this value of $y$ in $(2)$
$ 3 x+2\left[\frac{9-8 x}{5}\right]=4 $
Multiplying by $5 ,$
$ 15 x+18-16 x=20$
$15 x-16 x=20-18$
$-x=2$
$x=-2 $
From $(3)$
$\ y=\frac{9-8 x}{5}$
$=\frac{9-8(-2)}{5}$
$=\frac{25}{5}$
$=5$
$ y=5 $
View full question & answer→Question 794 Marks
Solve the following pairs of linear $($simultaneous$)$equation using method of elimination by substitution:$\frac{x}{6}+\frac{y}{15}=4, \frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
Answer$ \frac{x}{6}+\frac{y}{15}=4$
$ \Rightarrow \frac{5 x+2 y}{30}=4$
$ \Rightarrow 5 x+2 y=120$
$ \Rightarrow 5 x=120-2 y$
$\Rightarrow x=\frac{120-2 y}{5}\ldots .(1)$
And,
$\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
$\Rightarrow \frac{1}{3}\left(x-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{1}{3}\left(\frac{120-2 y}{5}-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{480-8 y-5 y}{20}=\frac{57}{4}$
$\Rightarrow \frac{480-13 y}{20}=\frac{57}{4}$
$\Rightarrow 480-13 y=285$
$\Rightarrow 13 y=195$
$ \Rightarrow y=15$
Substituting the value of $y$ in $(1),$ we have
$x=\frac{120-2 \times 15}{5}=\frac{120-30}{5}=\frac{90}{5}=18$
$\therefore$ Solution is $x=18$ and $y=15$
View full question & answer→Question 804 Marks
Solve the following pair of linear $($simultaneous$)$equation using method of elimination by substitution:$\frac{3 x}{2}-\frac{5 y}{3}+2=0,\frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$
Answer$ \frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$
$ \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$
$ \Rightarrow \frac{2 x+3 y}{6}=\frac{13}{6}$
$ \Rightarrow 2 x+3 y=13$
$ \Rightarrow 2 x=13-3 y$
$\Rightarrow x=\frac{13-3 y}{2}\ldots .(1)$
And,
$\frac{3 x}{2}-\frac{5 y}{3}+2=0$
$ \Rightarrow \frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{39-9 y}{4}-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{117-27 y-20 y}{12}=-2$
$ \Rightarrow \frac{117-47 y}{12}=-2$
$ \Rightarrow 117-47 y=-24$
$ \Rightarrow 47 y=141$
$ \Rightarrow y=3$
Substituting the value of $y$ in $(1),$ we have
$x=\frac{13-3 \times 3}{2}=\frac{13-9}{2}=\frac{4}{2}=2$
$\therefore$ Solution is $x=2$ and $y=3$.
View full question & answer→Question 814 Marks
Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12,\frac{3 x+2}{2}-\frac{4 y+3}{9}=13$
Answer$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12 \quad ($given$)$
$ \Rightarrow \frac{3(2 x+1)+7(5 y-3)}{21}=12$
$ \Rightarrow 6 x+3+35 y-21=252$
$ \Rightarrow 6 x+35 y-18=252$
$ \Rightarrow 6 x+35 y=270$
$ \Rightarrow 6 x=270-35 y$
$ \Rightarrow x=\frac{270-35 y}{6}$
$\frac{3 x+2}{2}-\frac{4 y+3}{9}=13 \quad($given$)$
$ \Rightarrow \frac{9(3 x+2)-2(4 y+3)}{18}=13$
$ \Rightarrow 27 x+18-8 y-6=234$
$ \Rightarrow 27 x-8 y+12=234$
$\Rightarrow 27 x-8 y=222\ldots .(1)$
Substituting $x=\frac{270-35 y}{6}$ in $(1)$, we get
$27\left(\frac{270-35 y}{6}\right)-8 y=222$
$ \Rightarrow 7290-945 y-48 y=1332$
$ \Rightarrow-993 y=-5958$
$ \Rightarrow y=6$
Substituting $y=6$ in $x=\frac{270-35 y}{6}$, we get
$x=\frac{270-35 \times 6}{6}=\frac{270-210}{6}=\frac{60}{6}=10$
$\therefore$ Solution is $x=10$ and $y=6$.
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