gibsonj338

\(x-6y=10\)  \(2x-2y=10\)

There are two ways to do it: by substitution and by subtraction

By substitution using the first equation first:

\(x-6y=10\)

\(x=10+6y\)

Substitute the \(x\) in the second equation with \(10+6y\)

\(2(10+6y)-2y=10\)

\(20+12y-2y=10\)

\(20+10y=10\)

\(10y=-10\)

\(y=-1\)

Substitute the \(y\) in the first equation with \(-1\)

\(x-6(-1)=10\)

\(x+6=10\)

\(x=4\)

\((-1,4)\)

OR

Substitute the \(y\) in the second equation with \(-1\)

\(2x-2(-1)=10\)

\(2x+2=10\)

\(2x=8\)

\(x=4\)

\((-1,4)\)

By substitution using the first equation second:  

\(x-6y=10\)

\(-6y=10-x\)

\(y=\frac{10-x}{-6}\)

Substitute the \(y\) in the second equation with \(\frac{10-x}{-6}\)

\(2x-2(\frac{10-x}{-6})=10\)

\(2x+\frac{10-x}{3}=10\)

\(\frac{10-x}{3}=10-2x\)

\(10-x=30-6x\)

\(-x=20-6x\)

\(-x+6x=20\)

\(5x=20\)

\(x=4\)

Substitute the \(x\) in the first equation with \(4\)

\(4-6y=10\)

\(-6y=6\)

\(y=-1\)

\((4,-1)\)

OR

Substitute the \(x\) in the second equation with \(4\)

\(2(4)-2y=10\)

\(8-2y=10\)

\(-2y=2\)

\(y=-1\)

\((4,-1)\)

By substitution using the second equation first:

\(2x-2y=10\)

\(2x=10+2y\)

\(x=\frac{10+2y}{2}\)

\(x=5+y\)

Substitute the \(x\) in the first equation with \(5+y\)

\(5+y-6y=10\)

\(5-5y=10\)

\(-5y=5\)

\(y=-1\)

Substitute the \(y\) in the second equation with \(-1\)

\(2x-2(-1)=10\)

\(2x+2=10\)

\(2x=8\)

\(x=4\)

\((4,-1)\)

OR

Substitute the \(y\) in the first equation with \(-1\)

\(x-6(-1)=10\)

\(x+6=10\)

\(x=4\)

\((4,-1)\)

By subtracton by subtracting the second equation from the first equation:

     \(x-6y=10\)

-  \(2x-2y=10\)

--------------------------

   \(-x-4y=0\)

\(-x=0+4y\)

\(-x=4y\)

\(x=-4y\)

Substitute the \(x\) in the first equation with \(-4y\)

\(-4y-6y=10\)

\(-10y=10\)

\(y=-1\)

Substitute the \(y\) in the second equation with \(-1\)

\(2x-2(-1)=10\)

\(2x+2=10\)

\(2x=8\)

\(x=4\)

\((4,-1)\)

OR

Substitute the \(x\) in the second equation with \(-4y\)

\(2(-4y)-2y=10\)

\(-8y-2y=10\)

\(-10y=10\)

\(y=-1\)

Substitute the \(y\) in the first equation with \(-1\)

\(x-6(-1)=10\)

\(x+6=10\)

\(x=4\)

\((4,-1)\)

By subtracton by subtracting the first equation from the second equation:

  \(2x-2y=10\)

-  \(x-6y=10\)

--------------------------

   \(x+4y=0\)

\(x=0-4y\)

\(x=-4y\)

Substitute the \(x\) in the first equation with \(-4y\)

\(-4y-6y=10\)

\(-10y=10\)

\(y=-1\)

Substitute the \(y\) in the second equation with \(-1\)

\(2x-2(-1)=10\)

\(2x+2=10\)

\(2x=8\)

\(x=4\)

\((4,-1)\)

OR

Substitute the \(x\) in the second equation with \(-4y\)

\(2(-4y)-2y=10\)

\(-8y-2y=10\)

\(-10y=10\)

\(y=-1\)

Substitute the \(y\) in the first equation with \(-1\)

\(x-6(-1)=10\)

\(x+6=10\)

\(x=4\)

\((4,-1)\)

\(2{sin}^{2}(x)+3sin(x) = -1\)

Let \(sin(x)=x\)

\(2{x}^{2}+3x=-1\)

\({2x}^{2}+3x+1=0\)

\((2x+1)(x+1)=0\)

\((2sin(x)+1)(sin(x)+1)=0\)

\(2sin(x)+1=0\)   \(sin(x)+1=0\)

\(2sin(x)+1=0\)

\(2sin(x)=-1\)

\(sin(x)=\frac{-1}{2}\)

\(sin(x)=-\frac{1}{2}\)

\({sin}^{-1}(sin(x))={sin}^{-1}(-\frac{1}{2})\)

\(x=-\frac{\pi}{6}\)

Because \(-\frac{\pi}{6}\) is not the only answer to the first half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers.  Represent this as \(x=-\frac{\pi}{6}+2\pi n\).  \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).

Because \({sin}^{-1}\)only gives one answer to the first half of the equation, to find the other answer, look at the unit circle and see where \(sin\) also equals \(-\frac{1}{2}\) and when you do that, you get that \(x=-\frac{5\pi}{6}.\)

Because \(-\frac{5\pi}{6}\)  is not the only answer to the first half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers.  Represent this as \(x=-\frac{5\pi}{6}+2\pi n\).  \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).

If having a negetive fraction at the begining of each answer looks weird, you can add \(2\pi\) to the negetive fraction to change the negetiive fraction to a positive fraction.  When you do that to the two answers to the first part of the equation, you get \(x=\frac{11\pi}{6}+2\pi n\) and \(x=\frac{7\pi}{6}+2\pi n\)

The two answers for the first half of the equation is \(x=-\frac{\pi}{6}+2\pi n\) \((x=\frac{11\pi}{6}+2\pi n)\) and \(x=-\frac{5\pi}{6}+2\pi n\) \((x=\frac{7\pi}{6}+2\pi n).\)

\(sin(x)+1=0\)

\(sin(x)=-1\)

\({sin}^{-1}(sin(x))={sin}^{-1}(-1)\)

\(x=-\frac{\pi}{2}\)

Because \(-\frac{\pi}{2}\) is not the only answer to the second half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers.  Represent this as \(x=-\frac{\pi}{2}+2\pi n\).  \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).

If you look at the unit circle, you will see that there is only one answer to the second half of the equation.  The only answer is \(x=-\frac{\pi}{2}+2\pi n.\)

If having a negetive fraction at the begining of the answer looks weird, you can add \(2\pi\) to the negetive fraction to change the negetiive fraction to a positive fraction.  When you do that to the answer to the first part of the equation, you get \(x=\frac{3\pi}{2}+2\pi n.\)

The answer to the second half of the equation is \(x=-\frac{\pi}{2}+2\pi n\) \((x=\frac{3\pi}{2}+2\pi n.).\) 

\({e}^{3x}=x\)

\({e}^{3x}≠x\)

\(ln({e}^{3x})≠ln(x)\)

\(3x\times ln(e)≠ln(x)\)

\(3x≠ln(x)\)

\(x≠\frac{ln(x)}{3}\)

\({3}^{-2}\times{a}^{4}\)

\(\frac{1}{{3}^{2}}\times{a}^{4}\)

\(\frac{1}{9}\times{a}^{4}\)

\(\frac{1}{9}{a}^{4}\)

\(\frac{{a}^{4}}{9}\)

\(\frac{{8x}^{-3}\times{y}^{2}\times{z}^{4}}{{5x}^{3}\times y\times{z}^{-2}}\)

\(\frac{8\times{y}^{2}\times{z}^{4}\times{z}^{2}}{{5x}^{3}\times y\times{x}^{3}}\)

\(\frac{8\times{y}^{2}\times{z}^{6}}{{5x}^{6}\times y}\)

\(\frac{8\times y\times{z}^{6}}{{5x}^{6}}\)

\(\frac{8y{z}^{6}}{{5x}^{6}}\)

\(365,537,274,738,959,995\times56,753,456,789,754,323,456,743,345,678\)

                                                                           \(2,924,298,197,911,679,960\)

                                                                         \(25,587,609,231,727,199,650\)

                                                                       \(219,322,364,843,375,997,000\)

                                                                   \(1,827,686,373,694,799,975,000\)

                                                                 \(14,621,490,989,558,399,800,000\)

                                                               \(109,661,182,421,687,998,500,000\)

                                                           \(1,096,611,824,216,879,985,000,000\)

                                                         \(14,621,490,989,558,399,800,000,000\)

                                                       \(255,876,092,317,271,996,500,000,000\)

                                                   \(2,193,223,648,433,759,970,000,000,000\)

                                                 \(18,276,863,736,947,999,750,000,000,000\)

                                               \(146,214,909,895,583,998,000,000,000,000\)

                                           \(1,096,611,824,216,879,985,000,000,000,000\)

                                           \(7,310,745,494,779,199,900,000,000,000,000\)

                                      \(109,661,182,421,687,998,500,000,000,000,000\)

                                  \(1,462,149,098,955,839,980,000,000,000,000,000\)

                                \(18,276,863,736,947,999,750,000,000,000,000,000\)

                              \(255,876,092,317,271,996,500,000,000,000,000,000\)

                          \(3,289,835,472,650,639,955,000,000,000,000,000,000\)

                        \(29,242,981,979,116,799,600,000,000,000,000,000,000\)

                      \(255,876,092,317,271,996,500,000,000,000,000,000,000\)

                  \(2,193,223,648,433,759,970,000,000,000,000,000,000,000\)

                \(18,276,863,736,947,999,750,000,000,000,000,000,000,000\)

              \(146,214,909,895,583,998,000,000,000,000,000,000,000,000\)

          \(1,096,611,824,216,879,985,000,000,000,000,000,000,000,000\) 

        \(18,276,863,736,947,999,750,000,000,000,000,000,000,000,000\)

      \(255,876,092,317,271,996,500,000,000,000,000,000,000,000,000\)

  \(2,193,223,648,433,759,970,000,000,000,000,000,000,000,000,000\)

\(18,276,863,736,947,999,750,000,000,000,000,000,000,000,000,000\)

Add the columns.

\(20,745,503,926,942,120,671,681,790,409,599,247,210,498,151,610\)

\(\frac{{(xy)}^{3}}{{xy}^{3}}\)

\(\frac{{x}^{3}{y}^{3}}{{xy}^{3}}\)

\({x}^{2}{y}^{0}\)

\({x}^{2}\times1\)

\({x}^{2}\)

\(\sqrt{81\times4}\)

\(\sqrt{324}\)

\(\sqrt{9\times9\times9\times9}\)

\(9\times9\)

\(18\)

If I understand you correctly, you want to change a positive number to a negative number.  There are several ways to do this.  You can add, subtract, multiply, divide, or take the negetive positive nth root of the positive number to change the positive number to a negetive number.

To change a positive number to a negetive number by adding:

Take a positive number and add a negetive number whose absolute value is greater than the absolute value of the positive number to get an answer of a negetive number.

\(5+(-6)=-1\)

To change a positive number to a negetive number by subtraction:

Take a positive number and subtract a positive number whose absolute value is greater than the absolute value of the first positive number to get an answer of a negative number.

\(5-6=-1\)

To change a positive number to a negative number by multiplication:

Take a positive number and multiply any negetive number to get an answer of a negetive number.

\(5\times(-6)=-30\)

To change a positive number to a negative number by division:

Take a positive number and divide any negetive number to get an answer of a negetive number.

\(5÷(-6)=-0.8333333333...\)

To change a positive number to a negative number by taking the negetive positive nth root of the positive number:

Take the positive number and take the positive nth root of the number then put the negetive sign in front of the positive number.

\(-\sqrt{4}=-2\)