Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $3 x+2 y+25=0$ $2 x+y+10=0$ Solution: GIVEN: $3 x+2 y+25=0$ $2 x+y+10=0$ To find: The solution of the systems of equation by the method of cross-multiplication: By cross multiplication method we get Also $y=\frac{20}{-1}$ Hence we get the value of $x=5$ and $y=-20$...

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In the below fig, lines AB and CD are parallel and P is any point as shown in the figure.

Question: In the below fig, lines AB and CD are parallel and P is any point as shown in the figure. Show that ABP + CDP = DPB. Solution: Given that AB ∥ CD Let EF be the parallel line to AB and CD which passes through P It can be seen from the figure Alternative angles are equal ABP = BPF Alternative angles are equal CDP = DPF ABP + CDP = BPF + DPF ABP + CDP = DPB Hence proved AB parallel to CD, P is any point To prove:ABP + BPD + CDP = 360 Construction: Draw EF ∥ AB passing through P Proof: Sin...

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Solve each of the following systems of equations by the method of cross-multiplication:

Question: Solve each of the following systems of equations by the method of cross-multiplication: $x+2 y+1=0$ $2 x-3 y-12=0$ Solution: GIVEN: $x+2 y+1=0$ $2 x-3 y-12=0$ To find: The solution of the systems of equation by the method of cross-multiplication: By cross multiplication method we get and $y=\frac{14}{-7}=-2$ Hence we get the value of $x=3$ and $y=-2$...

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Write the relation R =

Question: Write the relation R = {(x,x3) :xis a prime number less than 10} in roster form. Solution: R = {(x,x3) :xis a prime number less than 10} x= 2, 3, 5, 7 x3= 8, 27, 125, 343 R = {(2, 8), (3, 27), (5, 125), (7, 343)}...

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Prove that if the two arms of an angle are perpendicular to the two arms of another angle.

Question: Prove that if the two arms of an angle are perpendicular to the two arms of another angle. then the angles are either equal or supplementary. Solution: Consider be angles AOB and ACB Given 0A perpendicular to A0, also 0B perpendicular to BO To Prove:AOB + ACB = 180(or)AOB + ACB = 180 Proof: In a quadrilateral = A + O + B + C = 360 [Sum of angles of quadrilateral is 360] ⟹ 180 + O + B + C = 360 ⟹ O + C = 360 - 180 Hence AOB + ACB = 180 .... (1) Also, B + ACB = 180 ⟹ ACB = 180 - 90 = ACE...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $152 x-378 y=-74$ $-378 x+152 y=-604$ Solution: The given equations are: $152 x-378 y=-74 \ldots(i)$ $-378 x+152 y=-604 \ldots(i i)$ Multiply equation $(i)$ by 152 and equation $(i i)$ by 378 and add both equations we get Put the value of $x$ in equation $(i)$, we get $152 \times 2-378 y=-74$ $\Rightarrow-378 y=-378$ $\Rightarrow y=1$ Hence the value of $x=2$ and $y=1$...

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A = [1, 2, 3, 5] and B = [4, 6, 9].

Question: A = [1, 2, 3, 5] and B = [4, 6, 9]. Define a relation R from A to B by R = {(x,y) : the difference betweenxandyis odd,x A,y B}. Write R in Roster form. Solution: A = [1, 2, 3, 5] and B = [4, 6, 9] R = {(x,y) : the difference betweenxandyis odd,x A,y B} Forx= 1, $4-1=3$ and $6-1=5$ y= 4, 6 Forx= 2, $9-2=7$ y= 9 Forx= 3, $4-3=1$ and $6-3=3$ y= 4, 6 Forx= 5, $5-4=1$ and $6-5=1$ y= 4, 6 Thus, we have: R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $\frac{7 x-2 y}{x y}=5$ $\frac{8 x+7 y}{x y}=15$ Solution: The given equations are: $\frac{7 x-2 y}{x y}=5$ $7 x-2 y=5 x y$....(i) $\frac{8 x+7 y}{x y}=15$ $8 x+7 y=15 x y$$\ldots($ ii $)$ Multiply equation $(i)$ by 7 and equation $(i i)$ by 2 , add both equations we get Put the value of $y$ in equation $(i)$, we get $7 x-2 \times 1=5 x \times 1$ $\Rightarrow 2 x=2$ $\Rightarrow x=1$ Hence the value of $x=1$ and $y=1$...

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If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Question: If two straight lines are perpendicular to the same line, prove that they are parallel to each other. Solution: Given m perpendicular t and l perpendicular to t 1 = 2 = 90 Since, I and m are two lines and it is transversal and the corresponding angles are equal L ∥ M Hence proved...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ $\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$ Solution: The given equations are: $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ $\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$ Multiply equation $(i)$ by 3 and add both equations we get $\frac{6}{\sqrt{x}}+\frac{9}{\sqrt{y}}=6$ Put the value of $x$ in equation $(i)$, we get $\frac{2}{\sqrt{4}}+\frac{3}{\sqrt{y}}=2$ $\Rightarrow \sqrt{y}=3$ $\Rightarrow y=9$ Hence the ...

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In the below fig. if l ∥ m, n ∥ p and ∠1 = 85°. find ∠2.

Question: In the below fig. if l ∥ m, n ∥ p and 1 = 85. find 2. Solution: Corresponding angles are equal ⟹1 = 3 = 85 By using the property of co-interior angles are supplementary 2 + 3 = 180 2 + 55 = 180 2 = 180 85 2 = 95...

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Define a relation R on the set N of natural number by R

Question: Define a relation R on the set N of natural number by R = {(x,y) :y=x+ 5,xis a natural number less than 4,x,y N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range or R. Solution: R = {(x,y) :y=x+ 5,xis a natural number less than 4,x,y N} (i) ∵ x =1, 2, 3 y= 1 + 5, 2 + 5, 3 + 5 y = 6, 7, 8 Thus, we have: R = {(1, 6), (2, 7), (3, 8)}(ii) Now, Domain (R) = {1, 2, 3} Range (R) = {6, 7, 8}...

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In the below fig, state Which lines are parallel and why?

Question: In the below fig, state Which lines are parallel and why? Solution: Vertically opposite angles are equal ⟹EOC = DOK = 100 DOK = ACO = 100 Here two lines EK and CA cut by a third line and the corresponding angles to it are equal Therefore, EK ∥ AC....

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Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by

Question: Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by R = {(x,y) : 3xy= 0, wherex,y A}. Depict this relationship using an arrow diagram. Write down its domain, co-domain and range. Solution: A = [1, 2, 3,..., 14] R = {(x,y) : 3xy= 0, wherex,y A} Or, R = {(x,y) : 3x=y, wherex,y A} As $3 \times 1=3$ $3 \times 2=6$ $3 \times 3=9$ $3 \times 4=12$ Or, R = {(1, 3), (2, 6), (3, 9), (4, 12)} Domain (R) = {1, 2, 3, 4} Range (R) = {3, 6, 9, 12} Co-domain (R) = A...

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Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by

Question: Let A = [1, 2, 3, ......., 14]. Define a relation on a set A by R = {(x,y) : 3xy= 0, wherex,y A}. Depict this relationship using an arrow diagram. Write down its domain, co-domain and range. Solution: A = [1, 2, 3,..., 14] R = {(x,y) : 3xy= 0, wherex,y A} Or, R = {(x,y) : 3x=y, wherex,y A} As $3 \times 1=3$ $3 \times 2=6$ $3 \times 3=9$ $3 \times 4=12$ Or, R = {(1, 3), (2, 6), (3, 9), (4, 12)} Domain (R) = {1, 2, 3, 4} Range (R) = {3, 6, 9, 12} Co-domain (R) = A...

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In the below fig, PQ ∥ AB and PR ∥ BC.

Question: In the below fig, PQ ∥ AB and PR ∥ BC. If QPR = 102, determine ABC Give reasons. Solution: AB is produce to meet PR at K Since PQ ∥ AB QPR = BKR = 102 [corresponding angles] Since PR ∥ BC RKB + CBK = 180 [Since Corresponding angles are supplementary] CKB = 180 - 102 = 78 CKB = 78...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}$ $\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}$ Solution: The given equations are: $\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}$ $\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}$ Let $\frac{1}{3 x+y}=u$ and $\frac{1}{3 x-y}=v$ then equations are $u+v=\frac{3}{4} \ldots(i)$ $\frac{u}{2}-\frac{v}{2}=\frac{1}{8} \ldots$- (ii) Multiply equation (ii) by 2 and add both equations, we get $u+v=\frac...

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Let R be a relation from N to N defined by R =

Question: Let R be a relation from N to N defined by R = [(a,b) :a,b N anda=b2]. Are the following statements true? (i) (a,a) R for alla N (ii) (a,b) R ⇒ (b,a) R (iii) (a,b) R and (b,c) R ⇒ (a,c) R Solution: Given: R = [(a,b) :a,b N anda=b2] (i) (a,a) R for alla N. Here $2 \in \mathrm{N}$ hut? $\neq 2^{2}$ $\therefore(2,2) \notin \mathrm{R}$ False (ii) $(a, b) \in \mathrm{R} \Rightarrow(b, a) \in \mathrm{R}$ $\because 4=2^{2}$ $(4,2) \in \mathrm{R}$, but $(2,4) \notin \mathrm{R}$ False (iii) $(a...

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Prove

Question: $\frac{5 x-2}{1+2 x+3 x^{2}}$ Solution: Let $5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+B$ $\Rightarrow 5 x-2=A(2+6 x)+B$ Equating the coefficient ofxand constant term on both sides, we obtain $5=6 A \Rightarrow A=\frac{5}{6}$ $2 A+B=-2 \Rightarrow B=-\frac{11}{3}$ $\therefore 5 x-2=\frac{5}{6}(2+6 x)+\left(-\frac{11}{3}\right)$ $\Rightarrow \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x$ $=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 ...

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If below fig. if AB ∥ CD and CD ∥ EF, find ∠ACE.

Question: If below fig. if AB ∥ CD and CD ∥ EF, find ACE.. Solution: Since EF ∥ CD Therefore, EFC + ECD = 180 [co-interior angles are supplementary] ⟹ ECD = 180 - 130 = 50 Also BA ∥ CD ⟹ BAC = ACD = 70 [alternative angles] But, ACE + ECD = 70 ⟹ ACE = 70 - 50 = 20...

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In the below fig, show that AB ∥ EF

Question: In the below fig, show that AB ∥ EF Solution: Produce EF to intersect AC at K. Now,DCE + CEF = 35 + 145 = 180 Therefore, EF ∥ CD (Since Sum of Co-interior angles is 180) ... (1) Now,BAC = ACD = 57 ⟹ BA ∥ EF [Alternative angles are equal] ... (2) From (1) and (2) AB ∥ EF [Since, Lines parallel to the same line are parallel to each other] Hence proved....

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Let A = (x, y, z) and B = (a, b).

Question: Let A = (x,y,z) and B = (a,b). Find the total number of relations from A into B. Solution: Given: A= (x,y,z) andB= (a,b) Now, Number of elements in the Cartesian product of $A$ and $B=3 \times 2=6$ Number of relations from $A$ to $B=2^{6}=64$...

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Let A=

Question: Let $\mathrm{A}=\{a, b\}$. List all relations on $\mathrm{A}$ and find their number. Solution: Any relation inAcan be written as a set of ordered pairs. The only ordered pairs that can be included are(a,a), (a, b), (b, a) and (b, b).There are four ordered pairs in the set, and each subset is a unique combination of them. Each unique combination makesdifferent relations in A. { } [the empty set] {(a, a)} {(a, b)} {(a, a), (a, b)} {(b, a)} {(a, a), (b, a)} {(a, b), (b, a)} {(a, a), (a, b...

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In the below fig, AB ∥ CD ∥ EF and GH ∥ KL Find ∠HKL

Question: In the below fig, AB ∥ CD ∥ EF and GH ∥ KL Find HKL Solution: Produce LK to meet GF at N. Now, alternative angles are equal CHG = HGN = 60 HGN = KNF = 60 [Corresponding angles] Hence,KNG = 180 60 = 120 ⟹GNK = AKL = 120 [Corresponding angles] AKH = KHD = 25 [alternative angles] Therefore, HKL = AKH + AKL = 25 + 120 = 145...

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Determine the domain and range of the following relations:

Question: Determine the domain and range of the following relations: (i) $R=\{(a, b): a \in N, a5, b=4\}$ (ii) $S=\{(a, b): b=|a-1|, a \in Z$ and $|a| \leq 3\}$ Solution: (i)R= {(a,b) :a N,a 5,b= 4} We have: a= 1, 2, 3, 4 b= 4 R= {(1, 4), (2, 4), (3, 4), (4, 4)} Domain (R) = {1, 2, 3, 4} Range (R) = {4} (ii) $S=\{(a, b): b=|a-1|, a \in Z$ and $|a| \leq 3\}$ Now, $a=-3,-2,-1,0,1,2,3$ $b=|-3-1|=4$ $b=|-2-1|=3$ $b=|-1-1|=2$ $b=|0-1|=1$ $b=|1-1|=0$ $b=|2-1|=1$ $b=|3-1|=2$ Thus, we have: b= 4, 3, 2, ...

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