2/x+3/y=13
5/x-4/y=-2
Solve the equations by elimination and verify it by cross multiplication.....
Plz ans as soon as possible..
Answers
Answer:
2/x + 3/y = 13
5/x - 4/y = - 2
Answer
Putting 1/x = a and 1/y = b are reduce to
2a + 3b = 13 equation 1
5a - 4b = - 2 equation 2
By the elimination method
equation 1 is multiplied by 5 and equation 2 is multiplied by 2 are as follows ___
10a + 15b = 65 equation 3
10a - 8b = -4 equation 4
Now
By subtracting the equation 4 from equation 3 , we get
(10a-10a) + (15b + 8b) = 65 + 4
= 0 + 23b = 69
= b = 69/23 = 3
Now
putting b=3 in the equation
3 we get
10a + 15 × 3 = 65
10a + 45 = 65
10a = 65-45
a = 20/10 = 2
The reqd. solution is
since a=2 and b=3
Therefore we get
1/x=a and 1/y=b
So
NOW
1/x = 2 and 1/y = 3
2x = 1 and 3y = 1
x = 1/2 and y = 1/3
Hope this answer give you more knowledge
Step-by-step explanation:
Given,
2/x + 3/y = 13 ----(1)
5/x - 4/y = -2 ------(2)
Let 1/x = a and 1/y = b , We get
2a+3b=13 -----(3)
5a-4b = -2 -----(4)
On solving (3) * 4 & (4) * 3, we get
8a + 12b = 52
15a - 12b = -6
----------------------
23a = 46
=> a = 2
Substitute the value of a in equation (3), we get
=> 4 + 3b = 13
=> 3b = 13 - 4
=> 3b = 9
=> b = 9/3
=> b = 3
But 1/x = a = 2 => x = 1/2
1/y = b = 3 => y = 1/3
Thus,
x = (1/2), y = (1/3)
Verification :
Formula is :
[x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)]
On transposition, we get
2a + 3b - 13 = 0
5a - 4b + 2 = 0
By cross-multiplication method:
x/(3 * 2 - (-4)(-13)) = y/(-13 * 5 - 2 * 2) = 1/(2 * -4 - 5 * 3)
=> -x/46 = -y/69 = 1/-23
=> -x/46 = -1/23 and -y/69 = -1/23
=> x = 1/2, y = 1/3
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