Equations to solve

branko917

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Can somebody help me to solve this equation problem, i need expressions for y and z. I need all the steps in detail.

Equations.jpg

Thanks.
 
First step: Multiply both sides of equation 1 by 4.

You can now get rid of the nasty square root, and find an expression for y in terms of 4-z.

Second step: Substitute 4-z in the radical, and you have an expression in terms of y only to solve. Solve for y.

Third step: Substitute your value for y with the expression in the first step to determine z.
 
Just because I don't like big numbers, the first thing I would do is divide both sides of the first equation by 100 to get
\(\displaystyle 14y= \sqrt{25+ y^2+ (4- z)^2}\)
ans divide both sides of the second equation by 400 to get
\(\displaystyle 7.5(4-z)= \sqrt{25+ y^2+ (4- z)^2}\).

Although it is not necessary, I think I would replace "4- z" with "x" just to simplify it. That gives
\(\displaystyle 14y= \sqrt{25+ y^2+ x^2}\) and
\(\displaystyle 7.5x= \sqrt{25+ y^2+ x^2}\).

Do you see that the right sides of both equations are the same?
So \(\displaystyle 14y= 7.5x\) which tells us that x= (140/75)y and both equation are equivlent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).

And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
 
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Just because I don't like big numbers, the first thing I would do is divide both sides of the first equation by 100 to get
\(\displaystyle 14y= \sqrt{25+ y^2+ (4- z)^2}\)
ans divide both sides of the second equation by 400 to get
\(\displaystyle 7.5(4-z)= \sqrt{25+ y^2+ (4- z)^2}\).

Although it is not necessary, I think I would replace "4- z" with "x" just to simplify it. That gives
\(\displaystyle 14y= \sqrt{25+ y^2+ x^2}\) and
\(\displaystyle 7.5x= \sqrt{25+ y^2+ x^2}\).

Do you see that the right sides of both equations are the same?
So \(\displaystyle 14y= 7.5x\) which tells us that x= (140/75)y and both equation are equivlent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).

And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
Excuse me but 7/2 is not 7.5.
 
Just because I don't like big numbers, the first thing I would do is divide both sides of the first equation by 100 to get
\(\displaystyle 14y= \sqrt{25+ y^2+ (4- z)^2}\)
ans divide both sides of the second equation by 400 to get
\(\displaystyle 7.5(4-z)= \sqrt{25+ y^2+ (4- z)^2}\).

Although it is not necessary, I think I would replace "4- z" with "x" just to simplify it. That gives
\(\displaystyle 14y= \sqrt{25+ y^2+ x^2}\) and
\(\displaystyle 7.5x= \sqrt{25+ y^2+ x^2}\).

Do you see that the right sides of both equations are the same?
So \(\displaystyle 14y= 7.5x\) which tells us that x= (140/75)y and both equation are equivalent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).

And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).

But don't you make big numbers (as in $s) working here?
 
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\)
That works with the first equation, Halls, but not the second.

I've got one solution (so far), but I haven't finished checking other candidates.

?
 
Just because I don't like big numbers, the first thing I would do is divide both sides of the first equation by 100 to get
\(\displaystyle 14y= \sqrt{25+ y^2+ (4- z)^2}\)
ans divide both sides of the second equation by 400 to get
\(\displaystyle 7.5(4-z)= \sqrt{25+ y^2+ (4- z)^2}\).

Although it is not necessary, I think I would replace "4- z" with "x" just to simplify it. That gives
\(\displaystyle 14y= \sqrt{25+ y^2+ x^2}\) and
\(\displaystyle 7.5x= \sqrt{25+ y^2+ x^2}\).

Do you see that the right sides of both equations are the same?
So \(\displaystyle 14y= 7.5x\) which tells us that x= (140/75)y and both equation are equivlent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).

And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
Thanks a lot man, i needed this to solve one little problem in statics and i am a little rusty in math.
 
Just because I don't like big numbers, the first thing I would do is divide both sides of the first equation by 100 to get
\(\displaystyle 14y= \sqrt{25+ y^2+ (4- z)^2}\)
ans divide both sides of the second equation by 400 to get
\(\displaystyle 7.5(4-z)= \sqrt{25+ y^2+ (4- z)^2}\).

Although it is not necessary, I think I would replace "4- z" with "x" just to simplify it. That gives
\(\displaystyle 14y= \sqrt{25+ y^2+ x^2}\) and
\(\displaystyle 7.5x= \sqrt{25+ y^2+ x^2}\).

Do you see that the right sides of both equations are the same?
So \(\displaystyle 14y= 7.5x\) which tells us that x= (140/75)y and both equation are equivlent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).

And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
Please refrain from publishing "total answer" without significant effort from the OP. Suggest waiting at least 48 hrs before turnkey solutions are published.
 
I'd done something similar to what Halls did. I changed z=x, instead of 4-z=x. I then solved each equation for y.

I'd also graphed, to see what looked like intersecting hyperbolas. I was expecting a solution in each quadrant, but they seemed to have created an exercise that resists easy graphical investigation. The hyperbolas have been scaled and shifted, and maybe by scaling linear terms they created domain reductions just right for the branches to mostly pass between each other in the gaps, except for the one intersection point in QI. It's also interesting how they took a linear relationship between x and y, as well as the inverse relationship, multiplying one side by a radical and the other side by linear terms. I'm thinking that's why using approaches where the radical gets cancelled at the beginning leads to a false identity.

It's interesting also (to me) that there would be more solutions, if we considered the absolute value of the equations.

I checked Halls posted work, and, with the correction 7/2 = 3.5, his result matches mine. Since he posted a worked solution, I'll post the corrected answers.

y = 5/179 ∙ √179

z = 4 - 20/179 ∙ √179

?
 
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