Separate the constants and variables from $a b, 10,-2 x, \frac{-1}{3}$ and $\frac{x z}{2}$
Given :
The given terms are $a b, 10,-2 x, \frac{-1}{3}$ and $\frac{x z}{2}$.
To do :
We have to separate the constants and variables from the given terms.
Solution :
An Algebraic expression is one or a group of terms and may include variables, constants, operators and symbols.
For example, $3x + 2$ is an expression.
"x" is used in place of a value we don't know and is called "variable".
Therefore, from the given terms,
a and b are variables.
10 is constant.
$-2$ is a constant and x is a variable.
$\frac{-1}{3}$ is a constant.
$\frac{1}{2}$ is a constant and x,z are variables.
- Related Articles
- Subtract:(i) $-5xy$ from $12xy$(ii) $2a^2$ from $-7a^2$(iii) \( 2 a-b \) from \( 3 a-5 b \)(iv) \( 2 x^{3}-4 x^{2}+3 x+5 \) from \( 4 x^{3}+x^{2}+x+6 \)(v) \( \frac{2}{3} y^{3}-\frac{2}{7} y^{2}-5 \) from \( \frac{1}{3} y^{3}+\frac{5}{7} y^{2}+y-2 \)(vi) \( \frac{3}{2} x-\frac{5}{4} y-\frac{7}{2} z \) from \( \frac{2}{3} x+\frac{3}{2} y-\frac{4}{3} z \)(vii) \( x^{2} y-\frac{4}{5} x y^{2}+\frac{4}{3} x y \) from \( \frac{2}{3} x^{2} y+\frac{3}{2} x y^{2}- \) \( \frac{1}{3} x y \)(viii) \( \frac{a b}{7}-\frac{35}{3} b c+\frac{6}{5} a c \) from \( \frac{3}{5} b c-\frac{4}{5} a c \)
- Verify associativity of addition of rational numbers i.e., $(x + y) + z = x + (y + z)$, when:(i) \( x=\frac{1}{2}, y=\frac{2}{3}, z=-\frac{1}{5} \)(ii) \( x=\frac{-2}{5}, y=\frac{4}{3}, z=\frac{-7}{10} \)(iii) \( x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22} \)(iv) \( x=-2, y=\frac{3}{5}, z=\frac{-4}{3} \)
- Take away:(i) \( \frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x \) from \( \frac{x^{3}}{3}-\frac{5}{2} x^{2}+ \) \( \frac{3}{5} x+\frac{1}{4} \)(ii) \( \frac{5 a^{2}}{2}+\frac{3 a^{3}}{2}+\frac{a}{3}-\frac{6}{5} \) from \( \frac{1}{3} a^{3}-\frac{3}{4} a^{2}- \) \( \frac{5}{2} \)(iii) \( \frac{7}{4} x^{3}+\frac{3}{5} x^{2}+\frac{1}{2} x+\frac{9}{2} \) from \( \frac{7}{2}-\frac{x}{3}- \) \( \frac{x^{2}}{5} \)(iv) \( \frac{y^{3}}{3}+\frac{7}{3} y^{2}+\frac{1}{2} y+\frac{1}{2} \) from \( \frac{1}{3}-\frac{5}{3} y^{2} \)(v) \( \frac{2}{3} a c-\frac{5}{7} a b+\frac{2}{3} b c \) from \( \frac{3}{2} a b-\frac{7}{4} a c- \) \( \frac{5}{6} b c \)
- Find the product of $(-3 x y z)(\frac{4}{9} x^{2} z)(-\frac{27}{2} x y^{2} z)$ and verify the result for ; $x=2, y=3$ and $z=-1$
- If \( 2^{x}=3^{y}=12^{z} \), show that \( \frac{1}{z}=\frac{1}{y}+\frac{2}{x} \).
- If \( x+\frac{1}{x}=3 \), calculate \( x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}} \) and \( x^{4}+\frac{1}{x^{4}} \).
- Solve the following:If $x^{2}+\frac{1}{x^{2}}=3,$ find a) $ x-\frac{1}{x}$b) $x+\frac{1}{x} $
- If \( x^{4}+\frac{1}{x^{4}}=194 \), find \( x^{3}+\frac{1}{x^{3}}, x^{2}+\frac{1}{x^{2}} \) and \( x+\frac{1}{x} \)
- $\frac{x-1}{2}+\frac{2 x-1}{4}=\frac{x-1}{3}-\frac{2 x-1}{6}$.
- Solve for $x$:$\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}, x≠0, \frac{3}{2}, 2$
- Verify: $x\times(y\times z)=(x\times y)\times z$, where $x=\frac{1}{2},\ y=\frac{1}{3}$ and $z=\frac{1}{4}$.
- Verify the property \( x \times(y+z)=(x \times y)+(x \times z) \) for the given values of \( x,\ y \) and \( z \).\( x=\frac{-5}{2}, y=\frac{1}{2} \) and \( z=-\frac{10}{7} \)>
- Solve for x:$\frac{1}{( x-1)( x-2)} +\frac{1}{( x-2)( x-3)} =\frac{2}{3} \ ,\ x\neq 1,2,3$
- Which one of the following is a polynomial?(A) $\frac{x^{2}}{2}-\frac{2}{x^{2}}$(B) $\sqrt{2 x}-1$(C) $ x^{2}+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
- 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) \( \frac{7}{2} x^{3}-\frac{1}{2} x^{2}+\frac{5}{3}, \frac{3}{2} x^{3}+\frac{7}{4} x^{2}-x+\frac{1}{3} \) \( \frac{3}{2} x^{2}-\frac{5}{2} x-2 \)
Kickstart Your Career
Get certified by completing the course
Get Started