Given that
(2/x)+(3/y) = 13--------(1)
(5/x)-(4/y) = -2 -------(2)
Put 1/x = a and 1/y = b then
2a + 3b = 13 ----------(3)
On multiplying with 5 then
10a +15 b = 65 -------(4)
and
5a -4b= -2 ----------(5)
On multiplying with 2 then
10 a - 8b = -4 -------(6)
On Subtracting (6) from (4) then
10a + 15b = 65
10a - 8b = -4
(-)
_____________
0 + 23 b = 69
______________
⇛ 23b = 69
⇛ b = 69/23
⇛ b =3
On Substituting the value of b in (5)
5a -4b= -2
⇛ 5a -4(3) = -2
⇛ 5a -12 = -2
⇛ 5a = -2+12
⇛ 5a = 10
⇛ a = 10/5
⇛ a = 2
Now we have
a = 2
⇛1/x = 2
⇛ x = 1/2
and
b = 3
⇛1/y = 3
⇛ y = 1/3
<u>Answer :-</u>The solution for the given problem is (1/2,1/3)
<u>Check</u>: If x = 1/2 and y = 1/3 then
LHS = (2/x)+(3/y)
= 2/(1/2)+3/(1/3)
= (2×2)+(3×3)
= 4+9
= 13
= RHS
LHS=RHS is true
and
LHS=(5/x)-(4/y)
⇛ 5/(1/2)- 4/(1/3)
⇛(5×2)-(4×3)
⇛ 10-12
⇛ -2
⇛RHS
LHS = RHS is true
Answer:
This relationship is a function
Step-by-step explanation:
While a function may not have two output values (y) assigned to the same input value (x), it may have two input values (x) assigned to the output value (y).
The table follows that rule
Answer:
For this question there appears to be absolutely no association. The points are all over the place and there is not consistent factors at play here.
Answer:
Option b is correct (8,13).
Step-by-step explanation:
7x - 4y = 4
10x - 6y =2
it can be represented in matrix form as![\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-4%5C%5C10%26-6%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
A= ![\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-4%5C%5C10%26-6%5Cend%7Barray%7D%5Cright%5D%20)
X= ![\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
B= ![\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
i.e, AX=B
or X= A⁻¹ B
A⁻¹ = 1/|A| * Adj A
determinant of A = |A|= (7*-6) - (-4*10)
= (-42)-(-40)
= (-42) + 40 = -2
so, |A| = -2
Adj A= ![\left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-6%264%5C%5C-10%267%5Cend%7Barray%7D%5Cright%5D%20)
A⁻¹ = / -2
A⁻¹ = ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20)
X= A⁻¹ B
X= ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%283%2A4%29%20%2B%20%28-2%2A2%29%5C%5C%285%2A4%29%20%2B%20%28-7%2F2%2A2%29%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}12-4\\20-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12-4%5C%5C20-7%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}8\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
x= 8, y= 13
solution set= (8,13).
Option b is correct.
