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 #3
avatar+1904 
+5

If you mean \(\sqrt{x-y}\), you cannot seperate \(x\) and \(y\) into \(\sqrt{x}-\sqrt{y}\). If \(x\) and \(y\) was \(\sqrt{x\times y}\), then you can seperate \(x\) and \(y\) into \(\sqrt{x}\times\sqrt{y}\).  If \(x\) and \(y\) was \(\sqrt\frac{{x}}{y}\), then you an seperate \(x\) and \(y\) into \(\frac{{\sqrt x}}{\sqrt y}\).  If you mean \(\sqrt{x}-y\), you cannot add a \(\sqrt{}\) to y, that would change the problem entirely.

 

Example:

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x-y}\)

 

\(\sqrt{3-2}\)

 

\(\sqrt{1}\)

 

\(1\)

 

The answer is \(1\)

 

Now lets do \(\sqrt{x}-\sqrt{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-\sqrt{y}\)

 

\(\sqrt{3}-\sqrt{2}\)

 

\(1.7320508075688773...-1.414213562373095...\)

 

\(0.3178372451957823...\)

 

The answer is \(0.3178372451957823...\)

 

\(1\) and \(0.3178372451957823...\) are not the same answer.

 

Now lets do \(\sqrt{x\times y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x\times y}\)

 

\(\sqrt{3\times 2}\)

 

\(\sqrt{6}\)

 

\(2.4494897427831781...\)

 

The answer is \(2.4494897427831781...\)

 

Now lets do \(\sqrt{x}\times\sqrt{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}\times\sqrt{y}\)

 

\(\sqrt{3}\times\sqrt{2}\)

 

\(1.7320508075688773... \times 1.414213562373095...\)

 

\(2.4494897427831781...\)

 

The answer is \(2.4494897427831781...\)

 

\(2.4494897427831781...\) and \(2.4494897427831781...\) are the same answer.

 

Now lets do \(\sqrt\frac{{x}}{y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt\frac{{x}}{y}\)

 

\(\sqrt\frac{{3}}{2}\)

 

\(1.224744871391589...\)

 

The answer is \(1.224744871391589...\)

 

Now lets do \(\frac{{\sqrt x}}{\sqrt y}\)

 

Let \(x=3\) and let \(y=2\)

 

\(\frac{{\sqrt x}}{\sqrt y}\)

 

\(\frac{{\sqrt 3}}{\sqrt 2}\)

 

\(\frac{1.7320508075688773...}{1.414213562373095...}\)

 

\(1.224744871391589...\)

 

The answer is \(1.224744871391589...\)

 

\(1.224744871391589...\) and \(1.224744871391589...\) is the same answer

 

Now lets do \(\sqrt{x}-y\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-y\)

 

\(\sqrt{3}-2\)

 

\(1.7320508075688773...-2\)

 

\(-0.2679491924311227...\)

 

The answer is \(-0.2679491924311227...\)

 

Now let do \(\sqrt{x}-\sqrt y\)

 

Let \(x=3\) and let \(y=2\)

 

\(\sqrt{x}-\sqrt y\)

 

\(\sqrt{3}-\sqrt 2\)

 

\(1.7320508075688773...-1.414213562373095...\)

 

\(0.3178372451957823...\)

 

The answer is \(0.3178372451957823...\)

 

\(-0.2679491924311227...\) and \(0.3178372451957823...\) are not the same answer

 

Hope this helps

Jan 12, 2016
 #4
avatar+8581 
0
Jan 12, 2016
 #1
avatar+1832 
+4
Jan 12, 2016
 #1
avatar
0
Jan 12, 2016
 #1
avatar
+5

Explain how to solve the follwoing system of equations. What is the solution to the system?

 

2x+2y+z=-5

3x+4y+2z=0

x+3y+2z=1

 

Solve the following system:
{2 x+2 y+z = -5 |     (equation 1)
x+3 y+2 z = 1 |     (equation 2)
3 x+4 y+2 z = 0 |     (equation 3)
Swap equation 1 with equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
x+3 y+2 z = 1 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Subtract 1/3 × (equation 1) from equation 2:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+(5 y)/3+(4 z)/3 = 1 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Multiply equation 2 by 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
2 x+2 y+z = -5 |     (equation 3)
Subtract 2/3 × (equation 1) from equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x-(2 y)/3-z/3 = -5 |     (equation 3)
Multiply equation 3 by -3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+2 y+z = 15 |     (equation 3)
Subtract 2/5 × (equation 2) from equation 3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y-(3 z)/5 = 69/5 |     (equation 3)
Multiply equation 3 by 5/3:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y-z = 23 |     (equation 3)
Multiply equation 3 by -1:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+4 z = 3 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 4 × (equation 3) from equation 2:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+5 y+0 z = 95 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Divide equation 2 by 5:
{3 x+4 y+2 z = 0 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 4 × (equation 2) from equation 1:
{3 x+0 y+2 z = -76 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Subtract 2 × (equation 3) from equation 1:
{3 x+0 y+0 z = -30 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Divide equation 1 by 3:
{x+0 y+0 z = -10 |     (equation 1)
0 x+y+0 z = 19 |     (equation 2)
0 x+0 y+z = -23 |     (equation 3)
Collect results:
Answer: | {x = -10   y = 19    z = -23

Jan 12, 2016
 #2
avatar+8581 
0
Jan 12, 2016

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