6. Find two solutions for each of the following:
(i) 3x + 4y = 12 (ii) 3x + 5y =0 (iii) 4y + 5 = 0
Answers
Answered by
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Step-by-step explanation:
(i)3x+4y=12
=3x=12-4y
x=12-4y/3 by taking eq (x)
4y=12-3x
y=12-3x/4 by taking eq(y)
(ii)3x+5y=0
=3x=-5y 'x'
=5y=-3x 'y'
(iii)4y+5=0
4y=-5
tq
Answered by
0
Step-by-step explanation:
(1) 3x + 4y =12
by taking x=4 and y=0
3 × 4 + 4 × 0 = 12
12+0=12
12=12
L.H.S=R.H.S
(2) 3x + 5y = 0
by taking x = -5 and y = 3
3 × (-5) + 5×3 = 0
-15 + 15 = 0
0=0
L.H.S= R.H.S
(3) 4y + 5 = 0
by taking y = -5/4
4 × -5/4 + 5 = 0
AS , 4 and 4 gets cancelled out
-5 + 5 = 0
0=0
L.H.S=R.H.S
hope it helps pls mark as brainly
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