What is 14/5 as a mixed number ?
2 answers:
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The two linear equations in two variable is:
12 x + 3 y = 40
7 x - 4 y = 38
(a) For a system of equations in two Variable
a x + by = c
p x + q y = r
It will have unique solution , when
As, you can see that in the two equation Provided above
So, we can say the system of equation given here has unique solution.
(b). If point (2.5, -3.4) satisfies both the equations, then it will be solution of the system of equation, otherwise not.
1. 12 x+3 y=40
2. 7 x-4 y=38
Substituting , x= 2.5 , and y= -3.4 in equation (1) and (2),
L.H.S of Equation (1)= 1 2 × 2.5 + 3 × (-3.4)
= 30 -10.20
= 19.80≠ R.H.S that is 40.
Similarly, L H S of equation (2)= 7 × (2.5) - 4 × (-3.4)
= 17.5 +13.6
= 31.1≠R HS that is 38
So, you can Write with 100 % confidence that point (2.5, -3.4) is not a solution of this system of the equation.
<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>
<span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>