If you simply divide eqn(2) by eqn(1) - what do you get?Can somebody help me to solve this equation problem, i need expressions for y and z. I need all the steps in detail.
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Thanks.
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 equivlent to \(\displaystyle 14x= \sqrt{25+ 215x^2/75}\).
And because I don't like square roots- square both sides:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
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:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).
That works with the first equation, Halls, but not the second.\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\)
You mean setting up a proportion, right? I'd tried that, and it lead to y=z for all Real numbers (unless I goofed).If you simply divide eqn(2) by eqn(1) - what do you get?
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.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:
\(\displaystyle 196x^2= 25+ 215x^2/75\).
\(\displaystyle \frac{16125}{75}x^2= 25\).
\(\displaystyle x^2= \frac{75}{121}\)
\(\displaystyle y= 4- z= x= \frac{5}{11}\sqrt{3}\).