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3 years ago

Is the point (1,3) a solution to the linear equation 5x-9y=32? explain

Mathematics

2 answers:

murzikaleks [220]3 years ago

4 0

5x-9y=32
P (1,3)
Substitute:
5*(1)-9*(3)=32
5-27=32
-22=32 which is impossible.
Then the point (1,3) is not a solution.

harina [27]3 years ago

3 0

There are many ways to check if the point (1,3) is a solution to the linear equation $5x-9y=32$ .

Let us check it by expressing y in terms of x.

The given expression is 5x-9y=32. If we add -5x to both sides we will get:

$-9y=-5x+32$

Multiplying both sides by -1 we will get:

$9y=5x-32$

In order to isolate y, we will divide both sides by 9 to get:

$y=\frac{5}{9}x-\frac{32}{9}$

Now let us plug in the given value of x=1 from the point (1,3). This should give us y=3. Let us see if we get y=3 when we plug x=1 in the above equation.

$y=\frac{5}{9}\times 1-\frac{32}{9}=\frac{5-32}{9}$

$\therefore y=\frac{-27}{9}=-3$

Thus, we see that when x=1, y=-3 and that $y\neq 3$ and hence we conclude that the point (1,3) is not a solution to the original given linear equation 5x-9y=32.

For a better understanding of the explanation given here a graph has been attached. As can be seen from the graph, (1,3) does not lie on the straight line that represents 5x-9y=32, but (1,-3) does lie on it as we had just found out.

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$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

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