Solving Multiple Variable Equation with Exponents: ((Q_a)/(Q_d))^{1.85} * (P_d - P_e) + P_e = ...

[jzembrodt56]

I am struggling to solve for Qa in the following equation. Can anyone lend a hand? U was not sure how to enter N^1.85 so I apologize for the formatting.

[math](Q_a/Q_d )EXPONENT(1.85)×(P_d-P_e )+P_e = (Q_a/Q_f )EXPONENT(1.85)×(P_f-P_s )+P_s[/math]
These are equations for lines on a graph using an axis N^1.85 X-axis and I am trying to find the intersection so that I can plug the value for q back into my source equation. In my industry, we commonly find the solution by graphing the two lines. However, I would like to confirm the math behind it.

Thank you!
Josh Zembrodt

 

[stapel]

I am struggling to solve for Qa in the following equation. Can anyone lend a hand? U was not sure how to enter N^1.85 so I apologize for the formatting.

[math](Q_a/Q_d )EXPONENT(1.85)×(P_d-P_e )+P_e = (Q_a/Q_f )EXPONENT(1.85)×(P_f-P_s )+P_s[/math]

Is the equation meant to be as follows?

[imath]\qquad \left(\frac{Q_a}{Q_d}\right)^{1.85} \times \left(P_d - P_e\right) + P_e = \left(\frac{Q_a}{Q_f}\right)^{1.85} \times \left(P_f - P_s\right) + P_s[/imath]

 

[Dr.Peterson]

I am struggling to solve for Qa in the following equation. Can anyone lend a hand? I was not sure how to enter N^1.85 so I apologize for the formatting.

[math](Q_a/Q_d )EXPONENT(1.85)×(P_d-P_e )+P_e = (Q_a/Q_f )EXPONENT(1.85)×(P_f-P_s )+P_s[/math]
These are equations for lines on a graph using an axis N^1.85 X-axis and I am trying to find the intersection so that I can plug the value for q back into my source equation. In my industry, we commonly find the solution by graphing the two lines. However, I would like to confirm the math behind it.

Thank you!
Josh Zembrodt

You can type the equation as

(Q_a/Q_d)^{1.85}×(P_d-P_e )+P_e = (Q_a/Q_f )^{1.85}×(P_f-P_s )+P_s​

or put the same thing in the equation tool, which looks like this:

[math](Q_a/Q_d)^{1.85}×(P_d-P_e )+P_e = (Q_a/Q_f )^{1.85}×(P_f-P_s )+P_s[/math]
(There are fancier things you could do, but this is more than enough.)

But I'm not sure what you are asking for. This is a single equation, not a pair of equations that intersect.

But as for solving for [imath]Q_a[/imath], you can collect the terms containing that, and expand [math]\left(\frac{Q_a}{Q_d}\right)^{1.85}[/math] as [math]\frac{Q_a^{1.85}}{Q_d^{1.85}}[/math]
Then you can finish solving.

 

[jzembrodt56]

Is the equation meant to be as follows?

[imath]\qquad \left(\frac{Q_a}{Q_d}\right)^{1.85} \times \left(P_d - P_e\right) + P_e = \left(\frac{Q_a}{Q_f}\right)^{1.85} \times \left(P_f - P_s\right) + P_s[/imath]

Yes, that's it!

 

[jzembrodt56]

You can type the equation as

(Q_a/Q_d)^{1.85}×(P_d-P_e )+P_e = (Q_a/Q_f )^{1.85}×(P_f-P_s )+P_s​

or put the same thing in the equation tool, which looks like this:

[math](Q_a/Q_d)^{1.85}×(P_d-P_e )+P_e = (Q_a/Q_f )^{1.85}×(P_f-P_s )+P_s[/math]
(There are fancier things you could do, but this is more than enough.)

But I'm not sure what you are asking for. This is a single equation, not a pair of equations that intersect.

But as for solving for [imath]Q_a[/imath], you can collect the terms containing that, and expand [math]\left(\frac{Q_a}{Q_d}\right)^{1.85}[/math] as [math]\frac{Q_a^{1.85}}{Q_d^{1.85}}[/math]
Then you can finish solving.

Thank you Dr. Peterson.

Apologies for the confusion, what I showed was the first step of setting the equations equal to each other. The two source equations are shown below.

Each equation is in the plotted as a straight line on a graph where the x-axis = N^1.85, which is commonly used for water supply and fire protection analyses. Normal practice is to plot (Qd,Pd) and (0,Pe) for the first curve and (Qf,Pf), and (0,Ps) for the second one. Once plotted, we connect the points and find the intersection on the graph to determine (Qa,Pa). Plugging the equation into the Solve() function of my TI-89 and some online algebra tools were unable to solve it.

Thank you very much for your help!

 

[jzembrodt56]

Thank you Dr. Peterson.

Apologies for the confusion, what I showed was the first step of setting the equations equal to each other. The two source equations are shown below.

Each equation is in the plotted as a straight line on a graph where the x-axis = N^1.85, which is commonly used for water supply and fire protection analyses. Normal practice is to plot (Qd,Pd) and (0,Pe) for the first curve and (Qf,Pf), and (0,Ps) for the second one. Once plotted, we connect the points and find the intersection on the graph to determine (Qa,Pa). Plugging the equation into the Solve() function of my TI-89 and some online algebra tools were unable to solve it.
View attachment 36769
Thank you very much for your help!

I have attached an example graph that depicts the two equations. Note the x-axis is to the N^1.85.

I have also tried to write the equations as y=mx^1.85 + b using the points but think I am stuck there since one of my points for each line has x^1.85=0 or (q^1.85=0) in this case, which does not help in determining m; at least that I can figure out.

Thanks again for your insight!
Best regards,
Josh

 

[Dr.Peterson]

It's hard to follow explanations written in the jargon of a particular field. I think when you say "the x-axis is to the N^1.85", you mean that the distance on the x-axis is proportional to the 1.85th power of the value labeled; this fits what I see, because I observe that the ratio of distances for 2000 and 1000 is about 65/18=3.6, and 2^1.85=3.6.

So the idea is that you get a straight line in x and y if you let x=Q^1.85 and y=P. That's clear enough from the equations.

But what you want to do doesn't really depend on knowing this. You want to find the intersection of the two equations, which means solving for Q_a in the equation you gave us.

The work will be easier (whether by hand or by machine) if you change variables. Define

• [imath]x_a=Q_a^{1.85}[/imath]
• [imath]x_d=Q_d^{1.85}[/imath]
• [imath]x_f=Q_f^{1.85}[/imath]

Then the equations become [math]\frac{x_a}{x_d}(P_d-P_e )+P_e = \frac{x_a}{x_f}(P_f-P_s )+P_s[/math]
We want to solve for [imath]x_a[/imath].

To make the work a little easier, we can define a few more variables:

• [imath]A=P_d-P_e[/imath]
• [imath]B=P_f-P_s[/imath]
• [imath]C=P_s-P_e[/imath]

Multiplying everything by [imath]x_dx_f[/imath], we get [math]x_ax_fA+P_ex_dx_f = x_ax_d(P_f-P_s)+P_sx_dx_f[/math]
Collecting terms with [imath]x_a[/imath] on one side and the rest on the other, [math]x_ax_fA-x_ax_d(B) = P_sx_dx_f-P_ex_dx_f[/math]
Factoring out [imath]x_a[/imath], [math]x_a[x_fA-x_dB] = (P_s-P_e)x_dx_f[/math]
Dividing, [math]x_a=\frac{P_sx_dx_f-P_ex_dx_f}{x_fA-x_dB}=\frac{Cx_dx_f}{x_fA-x_dB}[/math]
Putting the original variables back, [math]Q_a^{1.85}=\frac{CQ_d^{1.85}Q_f^{1.85}}{Q_f^{1.85}A-Q_d^{1.85}B}[/math]
This can be simplified a little (fewer powers) as [math]Q_a^{1.85}=\frac{C}{\frac{A}{Q_f^{1.85}}-\frac{B}{Q_d^{1.85}}}[/math]
Finally, we can take a root: [math]Q_a=\left(\frac{C}{\frac{A}{Q_f^{1.85}}-\frac{B}{Q_d^{1.85}}}\right)^{1/1.85}[/math]
I hope you see that this is all basic algebra when you hide the complexity!

(But check it before you use it; I still may have made a mistake or two.)

 

[jzembrodt56]

It's hard to follow explanations written in the jargon of a particular field. I think when you say "the x-axis is to the N^1.85", you mean that the distance on the x-axis is proportional to the 1.85th power of the value labeled; this fits what I see, because I observe that the ratio of distances for 2000 and 1000 is about 65/18=3.6, and 2^1.85=3.6.

So the idea is that you get a straight line in x and y if you let x=Q^1.85 and y=P. That's clear enough from the equations.

But what you want to do doesn't really depend on knowing this. You want to find the intersection of the two equations, which means solving for Q_a in the equation you gave us.

The work will be easier (whether by hand or by machine) if you change variables. Define

• [imath]x_a=Q_a^{1.85}[/imath]
• [imath]x_d=Q_d^{1.85}[/imath]
• [imath]x_f=Q_f^{1.85}[/imath]

Then the equations become [math]\frac{x_a}{x_d}(P_d-P_e )+P_e = \frac{x_a}{x_f}(P_f-P_s )+P_s[/math]
We want to solve for [imath]x_a[/imath].

To make the work a little easier, we can define a few more variables:

• [imath]A=P_d-P_e[/imath]
• [imath]B=P_f-P_s[/imath]
• [imath]C=P_s-P_e[/imath]

Multiplying everything by [imath]x_dx_f[/imath], we get [math]x_ax_fA+P_ex_dx_f = x_ax_d(P_f-P_s)+P_sx_dx_f[/math]
Collecting terms with [imath]x_a[/imath] on one side and the rest on the other, [math]x_ax_fA-x_ax_d(B) = P_sx_dx_f-P_ex_dx_f[/math]
Factoring out [imath]x_a[/imath], [math]x_a[x_fA-x_dB] = (P_s-P_e)x_dx_f[/math]
Dividing, [math]x_a=\frac{P_sx_dx_f-P_ex_dx_f}{x_fA-x_dB}=\frac{Cx_dx_f}{x_fA-x_dB}[/math]
Putting the original variables back, [math]Q_a^{1.85}=\frac{CQ_d^{1.85}Q_f^{1.85}}{Q_f^{1.85}A-Q_d^{1.85}B}[/math]
This can be simplified a little (fewer powers) as [math]Q_a^{1.85}=\frac{C}{\frac{A}{Q_f^{1.85}}-\frac{B}{Q_d^{1.85}}}[/math]
Finally, we can take a root: [math]Q_a=\left(\frac{C}{\frac{A}{Q_f^{1.85}}-\frac{B}{Q_d^{1.85}}}\right)^{1/1.85}[/math]
I hope you see that this is all basic algebra when you hide the complexity!

(But check it before you use it; I still may have made a mistake or two.)

Thank you so much! I completely forgot about the ability to substitute complex variables for simple ones. Using this technique does reveal the equation to be basic algebra and allowed me to solve the equation (with a slight correction to what you have above). Thanks again for your help!