Page 38 - Algebra 1
P. 38

Explanation:
1. Here,
p1 =3,p2 =6,q1=k,q2 =–5,r1=7,r2 =–3 The two equations has no solution when
𝑝1 = π‘ž1 =ΜΈ π‘Ÿ1, then the pair of linear equation has no solution. 𝑝2 π‘ž2 π‘Ÿ2
3=π‘˜=ΜΈ7 6 βˆ’5 βˆ’3
3=π‘˜ 6 βˆ’5
1=π‘˜ 2 βˆ’5
βˆ’5 k=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.
 Let the numerator of the fraction is x
 The denominator of the fraction be y
π‘₯+2 = 1
π‘¦βˆ’3 3
3(x + 2) = (y – 3)
3x + 6 = y – 3
3x – y + 9 = 0......................(1)
π‘₯βˆ’3 = 1
𝑦+1 4
4(x – 3) = (y + 1)
4x – 12 = y + 1
4x – y – 13 = 0.......................(2)
Page 37 of 54
 ALGEBRA




































































   36   37   38   39   40