by Cramer rule please tell the answer
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The answer is given below :
RULE :
Let us consider two equations -
a1 x + b1 y = c1
a2 x + b2 y = c2
Then, the required solution be
x = (b2 c1 - b1 c2)/(a1 b2 - a2 b1)
and
y = (a1 c2 - a2 c1)/(a1 b2 - a2 b1)
SOLUTION :
The given equations are
x + 3y = 45 .....(i)
3x + 4y = 65 .....(ii)
So, the required solution by Cramer's Rule be
x = (45×4 - 65×3)/(1×4 - 3×3)
= (180 - 195)/(4 - 9)
= (-15)/(-5)
= 3
and
y = (1×65 - 3×45)/(1×4 - 3×3)
= (65 - 135)/(4 - 9)
= (-70)/(-5)
= 14
Therefore, the required solution be
x = 3 and y = 14.
VERIFICATION :
When x = 3, y = 14,
L.H.S. of (i)
= 3 + 3×14
= 3 + 42
= 45
= R.H.S. of (i)
and
L.H.S. of (ii)
= 3×3 + 4×14
= 9 + 56
= 65
= R.H.S. of (ii)
So, the values of x and y are verified.
Thank you for your question.
RULE :
Let us consider two equations -
a1 x + b1 y = c1
a2 x + b2 y = c2
Then, the required solution be
x = (b2 c1 - b1 c2)/(a1 b2 - a2 b1)
and
y = (a1 c2 - a2 c1)/(a1 b2 - a2 b1)
SOLUTION :
The given equations are
x + 3y = 45 .....(i)
3x + 4y = 65 .....(ii)
So, the required solution by Cramer's Rule be
x = (45×4 - 65×3)/(1×4 - 3×3)
= (180 - 195)/(4 - 9)
= (-15)/(-5)
= 3
and
y = (1×65 - 3×45)/(1×4 - 3×3)
= (65 - 135)/(4 - 9)
= (-70)/(-5)
= 14
Therefore, the required solution be
x = 3 and y = 14.
VERIFICATION :
When x = 3, y = 14,
L.H.S. of (i)
= 3 + 3×14
= 3 + 42
= 45
= R.H.S. of (i)
and
L.H.S. of (ii)
= 3×3 + 4×14
= 9 + 56
= 65
= R.H.S. of (ii)
So, the values of x and y are verified.
Thank you for your question.
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