plzzzzz someone help.
naira1902:
ohhh
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Answered by
17
Hi there!
Here's the answer:
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
From general form,
f(x) = ax²+bx+c= 0
g(x) = px²+qx+r= 0
Using cross multiplication method,
[ x / (br-ac) ]
= [ y / (cp-ra) ]
= [ 1 / (aq-bp) ]
---------(1)
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
Now,
given,
2x - 3y - 1.3 = 0
- x + y - 0.5 = 0
Comparing coefficients with general form,
a= 2 ; b= -3 ; c= -1.3
p= -1 ; q= 1 ; r= -0.5
Substituting values in (1), we get
x / [(-3×-0.5) - (1×-1.3)]
= y / [( -1.3 × -1) - (2×-0.5)]
= 1 / [(2×1) - (-3×-1)]
=> x/ (1.5+1.3) = y/ (1.3+1) = 1/ (2-3)
=> x/2.8 = y/2.3 = -1
•°• x= -2.8 & y= -2.3 are the required solutions.
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
©#£€®$
:)
Hope it helps
Here's the answer:
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
From general form,
f(x) = ax²+bx+c= 0
g(x) = px²+qx+r= 0
Using cross multiplication method,
[ x / (br-ac) ]
= [ y / (cp-ra) ]
= [ 1 / (aq-bp) ]
---------(1)
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
Now,
given,
2x - 3y - 1.3 = 0
- x + y - 0.5 = 0
Comparing coefficients with general form,
a= 2 ; b= -3 ; c= -1.3
p= -1 ; q= 1 ; r= -0.5
Substituting values in (1), we get
x / [(-3×-0.5) - (1×-1.3)]
= y / [( -1.3 × -1) - (2×-0.5)]
= 1 / [(2×1) - (-3×-1)]
=> x/ (1.5+1.3) = y/ (1.3+1) = 1/ (2-3)
=> x/2.8 = y/2.3 = -1
•°• x= -2.8 & y= -2.3 are the required solutions.
•°•°•°•°•°<><><<>><><>•°•°•°•°•°•
©#£€®$
:)
Hope it helps
Answered by
0
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