If X = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9} and Y = {3 , 6 , 9 , 12 , 15 , 18} , find $(X - Y)$ and $(Y -X)$ and illustrate them in Venn diagram.
Given :
X = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9} and Y = {3 , 6 , 9 , 12 , 15 , 18}.
To find :
We have to find $(X - Y)$ and $(Y -X)$.
Solution :
X = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9}
Y = {3 , 6 , 9 , 12 , 15 , 18}
$(X - Y)$ = {1 , 2 , 4 , 5 , 7 , 8 } [Values only in X but not in Y]
$(Y -X)$ = {12 , 15 , 18} [Values only in Y but not in X]
Venn diagram :
$(X - Y)$ = {1 , 2 , 4 , 5 , 7 , 8 }
$(Y -X)$ = {12 , 15 , 18}.
Related Articles
- Add the following algebraic expressions.a) \( x+5 \) and \( x+3 \)b) \( 3 x+4 \) and \( 4 x+9 \)c) \( 5 y-2 \) and \( 2 y+7 \) d) \( 8 y-3 \) and \( 5 y-6 \)
- The reduced form of $36 x^{2}-81 y^{2}$ isi$(6 x+9 y)(6 x-9 y) $ii$(6 x+9 y)(4 x-5) $iii.$(9 x+6 y)(9 x-6 y)$iv. $(9 y-6 x)(9 y+6 x) $
- Solve the following pairs of equations by reducing them to a pair of linear equations:(i) \( \frac{1}{2 x}+\frac{1}{3 y}=2 \)\( \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6} \)(ii) \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \)\( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)(iii) \( \frac{4}{x}+3 y=14 \)\( \frac{3}{x}-4 y=23 \)(iv) \( \frac{5}{x-1}+\frac{1}{y-2}=2 \)\( \frac{6}{x-1}-\frac{3}{y-2}=1 \)(v) \( \frac{7 x-2 y}{x y}=5 \)\( \frac{8 x+7 y}{x y}=15 \),b>(vi) \( 6 x+3 y=6 x y \)\( 2 x+4 y=5 x y \)4(vii) \( \frac{10}{x+y}+\frac{2}{x-y}=4 \)\( \frac{15}{x+y}-\frac{5}{x-y}=-2 \)(viii) \( \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \)\( \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \).
- If $x+y=5$ and $xy=6$ then find $x^3+y^3$.
- Verify the property: $x \times(y + z) = x \times y + x \times z$ by taking:(i) \( x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6} \)(ii) \( x=\frac{-12}{5}, y=\frac{-15}{4}, z=\frac{8}{3} \)(iii) \( x=\frac{-8}{3}, y=\frac{5}{6}, z=\frac{-13}{12} \)(iv) \( x=\frac{-3}{4}, y=\frac{-5}{2}, z=\frac{7}{6} \)
- Factorize:$4(x - y)^2 - 12(x -y) (x + y) + 9(x + y)^2$
- Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
- Factorize:\( x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3} \)
- For $x=\frac{3}{4} $and $y=\frac{-9}{8}$, insert a rational number between:$(x+y)^{-1} and x^{-1}+y^{-1} $
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- Add the following algebraic expressions(i) \( 3 a^{2} b,-4 a^{2} b, 9 a^{2} b \)(ii) \( \frac{2}{3} a, \frac{3}{5} a,-\frac{6}{5} a \)(iii) \( 4 x y^{2}-7 x^{2} y, 12 x^{2} y-6 x y^{2},-3 x^{2} y+5 x y^{2} \)(iv) \( \frac{3}{2} a-\frac{5}{4} b+\frac{2}{5} c, \frac{2}{3} a-\frac{7}{2} b+\frac{7}{2} c, \frac{5}{3} a+ \) \( \frac{5}{2} b-\frac{5}{4} c \)(v) \( \frac{11}{2} x y+\frac{12}{5} y+\frac{13}{7} x,-\frac{11}{2} y-\frac{12}{5} x-\frac{13}{7} x y \)(vi) \( \frac{7}{2} x^{3}-\frac{1}{2} x^{2}+\frac{5}{3}, \frac{3}{2} x^{3}+\frac{7}{4} x^{2}-x+\frac{1}{3} \) \( \frac{3}{2} x^{2}-\frac{5}{2} x-2 \)
- (i) \( x^{2}-3 x+5-\frac{1}{2}\left(3 x^{2}-5 x+7\right) \)(ii) \( [5-3 x+2 y-(2 x-y)]-(3 x-7 y+9) \)(iii) \( \frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+ \) \( \frac{1}{2} x y \)(iv) \( \left(\frac{1}{3} y^{2}-\frac{4}{7} y+11\right)-\left(\frac{1}{7} y-3+2 y^{2}\right)- \) \( \left(\frac{2}{7} y-\frac{2}{3} y^{2}+2\right) \)(v) \( -\frac{1}{2} a^{2} b^{2} c+\frac{1}{3} a b^{2} c-\frac{1}{4} a b c^{2}-\frac{1}{5} c b^{2} a^{2}+ \) \( \frac{1}{6} c b^{2} a+\frac{1}{7} c^{2} a b+\frac{1}{8} c a^{2} b \).
- If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.
- Verify the property: $x \times (y \times z) = (x \times y) \times z$ by taking:(i) \( x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9} \)(ii) \( x=0, y=\frac{-3}{5}, z=\frac{-9}{4} \)(iii) \( x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5} \)(iv) \( x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18} \)
- If $x – y = 7$ and $xy = 9$, find the value of $x^2+y^2$.
Kickstart Your Career
Get certified by completing the course
Get Started