Question

1. Without drawing the graph, find out whether the pair of linear equations 4x + 3y - 1 = 5 and 12x + 9y = 15 intersect at a point, are parallel or coincident?​

Answer 1

Answer:

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Step-by-step explanation:

4x + 3y - 1 = 5

4x +3y - 1 - 5 = 0

a1= 4 , b1 =3 , c1 =  - 6

12x + 9y = 15

12x + 9y - 15 = 0

a2 = 12 , b2= 9 , c2 = -15

a1  =  4  =   1

a2     12     3

b1  = 3  =   1

b2    9       3

c1  =  -6   =     6

c2     -15         15

a1    = b1     not equal to     c1

a2      b2                             c2

Therefore this pair of questions are co incident lines.

Answer 2

Answer:

Step-by-step explanation:

4x+3y-1=5              12x+9y=15

4x+3y=5+1

4x+3y=6

equations are 4x+3y=6  12x+9y=15

4x+3y=6

a1=4, b1=3 c1=6

12x+9y=15

a2=12, b2=9, c2=15

a1/a2=4/12=1/3

b1/b2=3/9=1/3

c1/c2=6/15=2/5

here a1/a2=b1/b2≠c1/c2

therefore it has a no solution

therefore the lines are pararell