Page 41 - Algebra

Explanation:
1. Here,
a1 = 3, a2 = 6, b1= k, b2 = –5, c1= 7, c2 = –3 The two equations have no solution when
𝑎1 = 𝑏1 ≠ 𝑐1, then the pair of linear equation has no solution. 𝑎2 𝑏2 𝑐2
3=𝑘≠7
6 −5 −3 3=𝑘
6 −5 1=𝑘
2 −5
5 2
2. Let the cost of one table be x Let the cost of one chair be y 4x + 3y = 800.............(1)
8x + 5y = 1500...........(2) Multiply (1) by 2
8x + 6y = 1600........... (3) (3) – (2)
y = 100
4x + 3(100) = 800
4x = 500
x = 125
Cost of 3 tables and 3 chairs = 3x + 3y = 3(125) + 3(100) = 375 + 300 = $675
3.
𝑥+2 = 1 𝑦−3 3
3(x + 2) = (y – 3)
3x + 6 = y – 3
3x – y + 9 = 0......................(1)
𝑥−3 = 1 𝑦+1 4
4(x – 3) = (y + 1)
4x – 12 = y + 1
4x – y – 13 = 0.......................(2)
Subtract the two equations. x – 22 = 0
x = 22
Put x = 22 in 𝑥+2 = 1 𝑦−3 3
22+2 = 1 𝑦−3 3
k =−
Let the numerator of the fraction be x
The denominator of the fraction be y
Page 40 of 177
Algebra I & II