Advertisements
Advertisements
प्रश्न
Subtract – 5a2 – 3a + 1 from the sum of 4a2 + 3 – 8a and 9a – 7.
योग
उत्तर
sum of 4a2 + 3 – 8a and 9a – 7
= 4a2 + 3 - 8a + 9a - 7 = 4a2 + a - 4
∴ (4a2 + a - 4) - (- 5a2 - 3a + 1)
= 4a2 + a - 4 + 5a2 + 3a - 1
= 4a2 + 5a2 + a + 3a - 4 - 1
= 9a2 + 4a - 5
shaalaa.com
Performs Operations (Addition and Subtraction) on Algebraic Expressions with Integral Coefficients Only.
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
If x = 4a2 + b2 - 6ab; y = 3b2 - 2a2 + 8ab and z = 6a2 + 8b2 - 6ab find: x - y - z
Simplify: 4(3x - 8) - 3(5x + 3) - 2(6x - 8)
Multiply: 4a + 5b and 4a - 5b
Multiply: - 3m2n + 5mn - 4mn2 and 6m2n
Copy and complete the following multi-plication:
6 - 3x + 2x2
× 1 + 5x - x2
Evaluate: (3x - 2y)(4x + 3y) (8x - 5y)
Evaluate: (a + 5)(3a - 2)(5a + 1)
Simplify: `"x"/2 -"x"/8`
Simplify: `"6k"/7 - ("8k"/9 - "k"/3)`
Simplify: `"3a"/8 + "4a"/5 - ("a"/2 + "2a"/5)`
