<span> 7x+2y=5;13x+14y=-1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
</span>System of Linear Equations entered :<span><span> [1] 7x + 2y = 5
</span><span> [2] 13x + 14y = -1
</span></span>Graphic Representation of the Equations :<span> 2y + 7x = 5 14y + 13x = -1
</span>Solve by Substitution :
// Solve equation [2] for the variable y
<span> [2] 14y = -13x - 1
[2] y = -13x/14 - 1/14</span>
// Plug this in for variable y in equation [1]
<span><span> [1] 7x + 2•(-13x/14-1/14) = 5
</span><span> [1] 36x/7 = 36/7
</span><span> [1] 36x = 36
</span></span>
// Solve equation [1] for the variable x
<span><span> [1] 36x = 36</span>
<span> [1] x = 1</span> </span>
// By now we know this much :
<span><span> x = 1</span>
<span> y = -13x/14-1/14</span></span>
<span>// Use the x value to solve for y
</span>
<span> y = -(13/14)(1)-1/14 = -1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
<span>
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Answer:
![\boxed{4 \sqrt[8]{ {d}^{3} } }](https://tex.z-dn.net/?f=%20%5Cboxed%7B4%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20%7D%20)
Step-by-step explanation:
![= > 4 {d}^{ \frac{3}{8} } \\ \\ = > 4({d}^{3 \times \frac{1}{8} }) \\ \\ = > 4( {d}^{3} \times {d}^{ \frac{1}{8} } ) \\ \\ = > 4( {d}^{3} \times \sqrt[8]{d} ) \\ \\ = > 4 \sqrt[8]{ {d}^{3} }](https://tex.z-dn.net/?f=%20%3D%20%20%3E%204%20%7Bd%7D%5E%7B%20%5Cfrac%7B3%7D%7B8%7D%20%7D%20%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%20%3E%204%28%7Bd%7D%5E%7B3%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B8%7D%20%7D%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%20%7Bd%7D%5E%7B%20%5Cfrac%7B1%7D%7B8%7D%20%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%5Csqrt%5B8%5D%7Bd%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%20%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20)
21/4 is equivalent to...
4 x 5 = 20
so, 21/4 = 5 1/4
5 1/4 = 5.25
5 1/4 = 5 2/8 = 42/8
42/8 = 84/16
Multiply the original DE by xy:
xy2(1+x2y4+1−−−−−−−√)dx+2x2ydy=0(1)
Let v=xy2, so that dv=y2dx+2xydy. Then (1) becomes
x(y2dx+2xydy)+xy2x2y4+1−−−−−−−√dxxdv+vv2+1−−−−−√dx=0=0
This final equation is easily recognized as separable:
dxxln|x|+CKxvKx2y2−1K2x4y4−2Kx2y2y2=−dvvv2+1−−−−−√=ln∣∣∣v2+1−−−−−√+1v∣∣∣=v2+1−−−−−√+1=x2y4+1−−−−−−−√=x2y4=2KK2x2−1integrate both sides
A ∩ B = {2, 4} [intersection (∩) means the common terms]