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Identify the following expression whether monomial, binomial, trinomial or quadrinomial expression: a b c + 1
Given a b c + 1
a b c + 1 is a binomial expression because it contains two terms
Write all the terms of the following algebraic expression:
2x + y³ − 3xy + 4y +3x +7
Given 2x + y³ − 3xy + 4y + 3x + 7
2x , y³, −3xy, 4y, 3x and 7 are the terms of the given algebraic expression.
Write the coefficient of y in the following: xyz
Given xyz
In the expression y is multiplied by xz
So The numerical coefficient of y is xz.
Write the coefficient of x2 in the following: \( – \left( {\dfrac{5}{2}} \right){\rm{ }}{\bf{a}}{{\bf{x}}^{\bf{2}}}\; + {\bf{2yx}}\)
Given:
is directly multiplied with
So, the numerical coefficient of is
Evaluate the following algebraic expression for x = 1, y = 1, z = 3, a = 2, b = 2, c = 1:
2ax + by + cz
Given x = 1, y = 1, z = 3, a = 2, b = 2, c = 1
Consider 2ax + by + cz
On substituting the given values
= 2(2) (1) + (2) (1) + (1) (3)
= –4 – 2 – 3
= –9
Add the following:
3abc, 5abc, 9abc, 7abc
Given 3abc, 5abc, 9abc, 7abc
Consider 3abc + (5abc) + (9abc) + (7abc)
= 3abc – 5abc + 9abc – 7abc
= (3 – 5 + 9 – 7) abc [by taking abc common]
= (12 – 12) abc
= 0abc = 0
Simplify the following:
2x²y³ +9y³x²
Given 2x²y³ +9y³x²
2x²y³ + 9y³x² = (2 + 9) x²y³
= 11x²y³
Add the following expression:
x3 3x2y + 2xy2– 4y3, 3×3– 5xy2 + 2x2y – 2y3
Given x³ –3x²y + 2xy²– 4y³, 3x³– 5xy² + 2x²y – 2y³
Collecting positive and negative like terms together, we get
= x³ +3x³ – 3x²y + 2x²y + 2xy² – 5xy² – 4y³– 2y³
= 4x³ – x²y – 3xy² – 6y³
Add the following expression:
4a – 6ab + 5b, –3a – ab – 8b and 4a + ab + 3b
Given 4a – 6ab + 5b, –3a – ab – 8b and 4a + ab + 3b
= (4a – 6ab + 5b) + (–3a – ab – 8b) + (4a + ab + 3b)
Collecting positive and negative like terms together, we get
= 4a – 3a + 4a – 6ab – ab + ab + 5b – 8b + 3b
= 5a – 7ab + ab + 8b – 8b
= 5a – 6ab
Add x2 – 2x + 2 to the sum of 3×2 – 2x and 2x + 7.
Given x² – 2x + 2, 3x² – 2x and 2x + 7
sum of 3x² – 2x and 2x + 7
= (3x² – 2x) + (2x +7)
=3x² – 2x + 2x + 7
= (3x² + 7)
Now, required expression = x² – 2x + 2+ (3x² + 7)
= x² + 3x² – 2x + 2 + 7
= 4x² – 2x + 9
Add 2×2 + 2xy + y2 to the sum of x2 – 3y2and 2×2 – y2 + 9.
Given 2×2 + 2xy + y2, x2 – 3y2and 2×2 – y2 + 9.
First, we have to find the sum of x2 – 3y2 and 2×2 – y2 + 9
= (x2 – 3y2) + (2×2 – y2 + 9)
= x2 + 2×2 – 3y2 – y2+ 9
= 3×2 – 4y2 + 9
Now, required expression = (2×2 + 2xy + y2) + (3×2 – 4y2 + 9)
= 2×2 + 3×2 + 2xy + y2 – 4y2 + 9
= 5×2 + 2xy – 3y2+ 9
Add a3+ b3 – 2 to the sum of a3 – 3b3 – 3ab + 7 and a3 + b3 + 3ab – 9.
Given a3+ b3 – 2, a3 – 3b3 – 3ab + 7 and a3 + b3 + 3ab – 9.
First, we need to find the sum of a3 – 3b3–3ab +7 and –a3 +b3 +3ab – 9.
= (a3 – 3b3– 3ab + 7) + ( a3 + b3 + 3ab – 9)
Collecting positive and negative like terms together, we get
= a3 – a3– 3b3+ b3 – 3ab + 3ab + 7 – 9
= –2b3 – 2
Now, the required expression = (a3 + b3 – 2) + (–2b3 – 2).
= a3+ b3– 2b3 – 2 – 2
= a3 – b3 – 4
Subtract: 3×3 −7×2 + 5x − 3 from 2 − 5x + 3×2 − 4×3
Given 3×3 −7×2 + 5x − 3 and 2 − 5x + 3×2 − 4×3
= (2 – 5x + 3×2 – 4×3) – (3×3 – 7×2 + 5x – 3)
= 2 – 5x + 3×2 – 4×3 – 3×3 + 7×2 – 5x + 3
= – 4×3– 3×3 + 7×2 + 3×2– 5x – 5x + 3 + 2
= – 7×3 + 10×2 – 10x +5
From 2p3 – 4 + 6p2, take away 10p2 − 6p3 + 2p − 6
Given 2p3 – 4 + 6p2, take away 10p2 − 6p3 + 2p − 6
= (2p3 – 4 + 6p2) – (10p2 – 6p3 + 2p – 6)
= 2p3 – 4 + 6p2 – 10p2 + 6p3 – 2p + 6
= 2p3 + 6p3 + 6p2 – 10p2– 2p – 4+ 6
= 8p3 – 4p2 – 2p + 2
From the sum of 3×2 − 5x + 1 and − 5×2 − 4x + 9 subtract 2×2 − 7x + 9.
First, we have to add 3×2 − 5x + 1 and − 5×2 − 4x + 9 then from the result we have to subtract 2×2 − 7x + 9.
= {(3×2 – 5x + 1) + ( 5×2 – 4x + 9)} – (2×2 – 7x + 9)
= {3×2 – 5x + 1 – 5×2 – 4x + 9} – (2×2 – 7x + 9)
= {3×2 – 5×2 – 5x – 4x + 1 + 9} – (2×2 – 7x + 9)
= { 2×2 – 9x +10} – (2×2 – 7x + 9)
= – 2×2 – 9x + 10 – 2×2 + 7x – 9
= – 2×2 – 2×2 – 9x + 7x + 10 – 9
= – 4×2 – 2x + 1
Subtract the sum of 13x – 2y + 7z and – 3z + 3x + 3y from the sum of 3x – 2y – 2z and 1x + 2y – 7.
First, we have to find the sum of 13x – 2y + 7z and – 3z + 3x + 3y
Therefore, sum of (13x – 2y + 7z) and (–3z + 3x + 3y)
= (13x – 2y + 7z) + (–3z + 3x + 3y)
= (13x – 2y + 7z – 3z + 3x + 3y)
= (13x + 3x – 2y + 3y + 7z – 3z)
= (16x + y + 4z)
Now we have to find the sum of (3x – 2y – 2z) and (1x + 2y – 7)
= (3x – 2y – 2z) + (1x + 2y – 7)
= (3x – 2y – 2z + 1x + 2y – 7)
= (3x + 1x – 2z – 7)
= (4x – 2z – 7)
Now, required expression = (4x – 2z – 7) – (16x + y + 4z)
= 4x – 2z – 7 – 16x – y – 4z
= 4x – 16x – y – 2z – 4z – 7
= –12x – y – 6z – 7
What should be added to 2xy – 6yz + 4zx to get 4xy – 6zx + 4yz + 14?
By subtracting 2xy – 6yz + 4zx from 4xy – 6zx + 4yz + 14, we get the required expression.
Therefore, required expression =(4xy–6zx+4yz+14)–(2xy–6yz+4zx)
= 4xy – 6zx + 4yz + 14 – 2xy + 6yz – 4zx
= 4xy – 2xy – 6zx – 4zx + 4yz + 6yz + 14
= 2xy – 10zx + 10yz + 14
How much is x – y + 3z greater than x + 5y – 7?
By subtracting x – y + 3z from x + 5y – 7 we can get the required expression,
Required expression = (x – y + 3z) – (x + 5y – 7)
= x – y + 3z – x – 5y + 7
Collecting positive and negative like terms together, we get
= x – x – y – 5y + 3z + 7
= – 6y + 3z + 7
How much does a2 − ab + 2b2 exceed 2a2 − 7ab + 5b2?
By subtracting 2a2 − 7ab + 5b2 from a2 − ab + 2b2 we get the required expression
Required expression = (a2– ab + 2b2) – (2a2 – 7ab + 5b2)
= a2– ab + 2b2 – 2a2 + 7ab – 5b2
Collecting positive and negative like terms together, we get
= a2 – 2a2 – ab + 7ab + 2b2 – 5b2
= – a2 + 6ab – 3b2
If P = 4×2 + 5xy − 7y2, Q = 4y2 − 2×2 − 3xy and R = −2×2 – 2xy + 3y2, then that P + Q + R is
Given P = 4×2 + 5xy − 7y2, Q = 4y2 − 2×2 − 3xy and R = −2×2 – 2xy + 3y2
Now we have to prove P + Q + R = 0,
Consider P + Q + R = (4×2+5xy–7y2) + (4y2 –2×2–3xy) + ( 2x²–2xy+3y2)
= 4×2 + 5xy – 7y2 + 4y2 – 2×2 – 3xy – 2×2 – 2xy + 3y2
Collecting positive and negative like terms together, we get
= 4×2– 2×2 – 2×2 + 5xy – 3xy – 2xy – 7y2 + 4y2 + 3y2
= 4×2– 4×2 + 5xy – 5xy – 7y2 + 7y2
= 0
If P = a2 − b2 + ab, Q = a2 + 2b2 − 3ab, R = b2 + b, S = a2 − 2ab and T = −a2 + b2 – ab + a. Find P + Q + R + S – T.
Given P = a2 − b2 + ab, Q = a2 + 2b2 − 3ab, R = b2 + b, S = a2 − 2ab and T = −a2 + b2 – ab + a.
Now we have to find P + Q + R + S – T
Substituting all values, we get
Consider P + Q + R + S – T = {(a2 – b2 + ab) + (a2 + 2b2 – 3ab) + (b2 + b) + (a2 – 2ab)} – (a2 + b2 – ab + a)
= {a2 – b2 + ab + a2 + 2b2 – 3ab + b2 + b + a2 – 2ab} ( a2 + b2 – ab + a)
= {3a2 + 2b2 – 4ab + b } – (a2 + b2 – ab + a)
= 3a2+ 2b2 – 4ab + b + a2 – b2 + ab – a
Collecting positive and negative like terms together, we get
3a2+ a2 + 2b2 – b2 – 4ab + ab – a + b
= 4a2 + b2– 3ab – a + b
If the value of the expression x² – 3x + k for x = 0 is 5, then the value of k is
Given expression: x²3x+k
Substitute x by 0 in the expression, we get 5:
= (0)²3(0)+k=5
⇒ k = 5
Find the value of the expression a³ + b³ + c³ – 3abc for a = 1, b = 2, c = 3
Given expression: a³ + b³ + c³ – 3abc
Substitute the values a=1, b=2 and c=3
= (1)³ + (2)³ + (3)³ – 3(1)(2)(3)
= 1 + 8 + 27 – 18 = 18
Find the value of the expression 3p + 8 for p = 2
Given expression: 3p+8
Substitute p with 2:
3(2) + 8 = 2
Find the value of the expression x³ – 2(x – 5) for x = 5
Given expression: x³ – 2(x –5)
Substitute x by 5
(5)³ – 2(5 – 5) = 125
Puja travelled 3x km distance by walk, 2y km by cycle and 9xy km by bus. The total distance covered by Rakhi in an algebraic expression is____.
Distance travelled by foot = 3x km
Distance travelled by cycle = 2y km
Distance travelled by bus = 9xy
Total distance travelled= distance by foot + cycle + bus
Substitute the values
Total distance = (3x+2y+9xy) km
So the algebraic expression will be 3x+2y+9xy km.
The equation for the statement; ‘Half of a number added to 7 is 13’ is______ .
Let the number be x
Half of x =
added to 7 = + 7
It’s given that half of number added to 7 = 13
So required expression: + 7 = 13
Akash had ₹200 with him. He gave ₹x to Ajay, ₹x/2 to Vidhu and is left with ₹ x/2. The amount he gave to Vidhu is_______.
Money Akash had = 200 ₹
Amount he gave to Ajay = x ₹
Amount he gave to Vidhu = ₹
Total amount given = amount given to Ajay + amount given to Vidhu
So Total amount given = x+₹
Finally he is left with amount =
So, 200 – {x+} =
200 – x – =
200 – x = +
200 – x = x
200 = 2x
x = 100
Now, amount given to vidhu =
Substitute the value of x
Amount given to Vidhu = = 50
Ages of two friends are in the ratio 3 : 2. If the sum of their ages is 55, then their ages are respectively __________.
Let their ages be 3x and 2x
Sum of their ages = 3x + 2x
Which is given to be 55
So, 3x + 2x = 55
5x = 55
x = = 11
So, the ages are 3x = 3 × 11 = 33
And, 2x = 2 × 11 = 22
Equation for the statement ‘Thrice the length (l) of a room added to 7 is 340 metres’ is ______.
Thrice the length (l) = 3l
When 7 added to 3l = 3l+7
This is given to be = 340
So, 3l + 7 = 340
If four times of a number is 48, then the number is______.
Let the number be x
Given:
4 times the number = 4x = 48
So, x = = 12
So, the required number is 12
An algebraic expression 2x – y can be written in statement as______.
Here x is a number which is taken twice i.e. 2x
Now y is subtracted from 2x i.e 2x7
So, the given expression is ‘y less than 2x’.
In 3(a – 1) + 4 = 7, the value of ‘a’ is______.
Given: 3(a – 1) + 4 = 7
3a – 3 + 4 = 7
3a + 1 = 7
3a = 7 – 1 = 6
a = = 2
Twice of a number is added to 18, then the sum is 46. The number is _______.
Let the number be x
Now twice of that number added to 18 = 2x + 18
This sum is given to be 46.
So, 2x + 18 = 46
2x = 46 – 18
2x = 28
x = = 14
So, the required number is 14
The method of finding solution by repeated attempts using varied inputs for a variable until success is called _____.
In trialanderror method, we use different values for a variable and check if we get the desired result or not. This is continued till we get success or all possible/required inputs are tried.
Value of x in \(\dfrac{{\rm{x}}}{2} + \dfrac{1}{2} = 4\)is _________.
Six less than a number equals to two. What is the number?
Let the number be ‘x’.
According to condition, we have x – 6 = 2
By inspections, we have 8 – 6 = 2
∴ x = 8
Thus, the required number is 8.
To get the value of ‘a’, the number to be divided with on either side of equation 7a = 35 is
Given equation: 7a = 35
To get the value of ‘a’ we have to get rid of ‘7’ from 7a.
So, we multiply both sides with 7
Which is,
a = 5
Think of a number, add 2 to it and then multiply the sum by 6, the result is 42.
Let the number be x.
∴ Sum of x and 2 = x + 2
Now by multiplying the sum by 6, we get
6 × (x + 2) = 42
⇒ 6 × x + 6 × 2 = 42
⇒ 6x + 12 = 42
By inspection, we get
6 × 5 + 12 = 42
⇒ 30 + 12 = 42
∴ 42 = 42
So, the required number = 5
f x = 2, y = 3 and 2 = 5, find the value of x – y + z
x – y + z = 2 – 3 + 5 = 1 + 5 = 4
The value of x in the equation 4(x + 10) = 140 is_________.
Given: 4(x + 10) = 140
4x + 4×10 = 140
4x = 140 – 40
4x = 100
x = = 25
Here is a pattern of houses with matchsticks:
Write the general rule for this pattern.
One house is made of 6 matchsticks i.e. 6 x 1
Two houses are made of 12 matchsticks i.e. 6 x 2
Three houses are made of 18 matchsticks i.e. 6 x 3
∴ Rule is 6n where n represents the number of houses.
If we multiply 5x and (– 4xyz), then we get:
The volume of a cuboid with length, breadth and height as 5x, 3×2 and 7×4 respectively is:
Volume of cuboid = Length × breadth × height
cubic units
The algebraic expression for the statement “Product of x and 3 added to the product of ‘ab’ and y’ is ______
Product of x and 3 = 3x
Product of ‘ab’ and y = aby
When both added together = 3x + aby
So, the required expression is 3x+aby
How many terms are there in the expression 7x² + 5x – 5 ?
Terms – 7x², 5x, 5.
A kangaroo can jump up to 10.5 m in a single jump. If the kangaroo jumped 105 m altogether, how many jumps did it make?
Distance covered by kangaroo in one jump = 10.5
Total distance covered = 105
Total number of jumps =
Total number of jumps = = 10
So, the kangaroo covers 105 m in 10 jumps.
If e = f, then e + x = _______ .
Given: e = f
Then e + x = ?
Substitute the value of e in the expression. Value of e = f.
So, e + x = f + x
The algebraic expression for the statement ‘x multiplied by itself and added to 5′ is____.
x multiplied by itself i.e., x = x × x = x²
Now x² added to 5 = x² + 5
So, the required algebraic expression is x² + 5
If a = b, then ax +3 =
Given: a=b
We have to find ax+3
Substitute a with b in the expression, we get:
ax + 3 = (b) x + 3 = bx + 3