\(\displaystyle 5x -2y = 25\)

\(\displaystyle 4x-2y=24\)

To solve this system of equations,  subtract the second equation from the first.

       \(\displaystyle 5x - 2y = 25\)

     \(\displaystyle -4x + 2y = -24\)

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                   \(\displaystyle x=1\)

Now, substitute in the value for x into one of the equations to solve for the value of \(\displaystyle y\).

\(\displaystyle \\5x-2y = 25 \\5(1)-2y=25 \\5-2y=25\)

Now subtract five from each side.

\(\displaystyle \\5-5-2y=25-5 \\-2y=20\)

Divide both sides by negative 2:

\(\displaystyle \\ \frac{-2y}{-2}=\frac{20}{-2} \\ \\y=-10\)

\(\displaystyle x = 1\)

\(\displaystyle (1,-10)\) is the correct answer.

To solve this system of equations, set both equations equal to one another.

\(\displaystyle \frac{1}{5}x +2 = -\frac{1}{5}x-4\)

Add \(\displaystyle \frac{1}{5}x\) to both sides of the equation.

\(\displaystyle \frac{1}{5}x + \frac{1}{5}x + 2= -\frac{1}{5}x +\frac{1}{5}x -4\)

\(\displaystyle \frac{2}{5}x + 2 = -4\)

Subtract \(\displaystyle 2\) from both sides of the equation.

\(\displaystyle \frac{2}{5}x + 2 -2 = -4-2\)

\(\displaystyle \frac{2}{5}x =-6\)

Multiply both sides of the equation by \(\displaystyle \frac{5}{2}\).

\(\displaystyle \frac{2}{5}x \times \frac{5}{2} = \frac{-6}{1} \times \frac{5}{2}\)

\(\displaystyle x = \frac{-30}{2}\)

\(\displaystyle x = -15\)

Plug the value of \(\displaystyle x\), which is \(\displaystyle -15\) into one of the equations to get the value of \(\displaystyle y.\)

\(\displaystyle -\frac{1}{5 }\times \frac{-15}{1 } - 4 = y\)

\(\displaystyle 3-4 = y\)

\(\displaystyle y = -1\)

\(\displaystyle (-15,-1)\)  is the correct answer for this system of equations. 

Step 1: Write the two equations, one below another and line up the terms.

\(\displaystyle x-y=5\)
\(\displaystyle x+y=7\)
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Step 2: We see that we have \(\displaystyle -y\) and \(\displaystyle y\). We can add these two equations up, which will isolate y and let us solve for x.

\(\displaystyle x{\color{Red} -y}=5\)
\(\displaystyle x+{\color{Blue} y}=7\) We add here.
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\(\displaystyle 2x=12\)

Step 3: We will isolate x by itself. We need to divide by 2 on both sides to get x.

\(\displaystyle \frac {2x}{2}=\frac {12}{2}\)
\(\displaystyle \rightarrow x=6\)

Step 4: We found x, so we can plug in that value into any one of the two equations and solve for y. Let's choose equation (1).

(1)...\(\displaystyle x-y=5\)
\(\displaystyle (6)-y=5\). Isolate y by itself. We are going to subtract 6 from both sides.
\(\displaystyle 6-6-y=5-6\). Simplify the left hand side.
\(\displaystyle -y=-1\)

Step 5: We will divide by -1 to get the value of y.

\(\displaystyle \frac {-y}{-1}=\frac {-1}{-1}\)
\(\displaystyle \rightarrow y=1\)

The values that solve this system of equations is \(\displaystyle x=6\) and \(\displaystyle y=1\).

To answer this question you must first solve for one of the variables. This can be done with either variable with either equation. In this example of how to solve the problem we will solve for y using the second equation

\(\displaystyle 8x+4y=12\)

subtract 8x from both sides

\(\displaystyle 4y=12-8x\)

divide both sides by y

\(\displaystyle y=3-2x\)

Now we plug this into the first equation for the y variable

\(\displaystyle 4x+3(3-2x)=11\)

Distribute the 3

\(\displaystyle 4x+9-6x=11\)

Simplify

\(\displaystyle -2x+9=11\)

subtract 9 from both sides

\(\displaystyle -2x=2\)

divide by -2 on both sides

\(\displaystyle x=-1\)

Using this we solve for y in the second equation

\(\displaystyle 8(-1) +4y=12\)

simplify

\(\displaystyle -8+4y=12\)

add 8 to both sides

\(\displaystyle 4y=20\)

divide by 4 on both sides

\(\displaystyle y=5\)

Final answer\(\displaystyle x=-1\) and \(\displaystyle y=5\)

In order to find the eigenvalues of a matrix, apply the following formula:

\(\displaystyle det(A-\lambda I) =0\)

\(\displaystyle I\) is the identity matrix.

\(\displaystyle \begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\\ -2&1-\lambda \end{bmatrix}\)

Compute the determinant and set it equal to zero.

\(\displaystyle (1-\lambda)(1-\lambda)-(2)(-2)=0\)

\(\displaystyle 1-2\lambda+\lambda^2+4=0\)

\(\displaystyle \lambda^2-2\lambda+5=0\)

Solve for lambda by using the quadratic formula.

\(\displaystyle \lambda_{1,2}=1\pm2i\)

In order to find the eigenvalues of a matrix, apply the following formula:

\(\displaystyle det(A-\lambda I) =0\)

\(\displaystyle I\) is the identity matrix.

\(\displaystyle \begin{bmatrix} 1&2 \\ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\\ -2&1-\lambda \end{bmatrix}\)

Compute the determinant and set it equal to zero.

\(\displaystyle (1-\lambda)(1-\lambda)-(2)(-2)=0\)

\(\displaystyle 1-2\lambda+\lambda^2+4=0\)

\(\displaystyle \lambda^2-2\lambda+5=0\)

Solve for lambda by using the quadratic formula.

\(\displaystyle \lambda_{1,2}=1\pm2i\)

The fifth degree Taylor polynomial approximating \(\displaystyle sin(x)\) centered at \(\displaystyle x=0\) is: 

\(\displaystyle sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}\)

The Lagrange error is the absolute value of the next term in the sequence, which is equal to \(\displaystyle \left |\frac{x^7}{7!} \right |\).

We need only evaluate this at \(\displaystyle x=1\) and thus we obtain \(\displaystyle \frac{1^7}{7!} = \frac{1}{7!}\)

We can show that the series \(\displaystyle \sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}\)  diverges using the ratio test.

\(\displaystyle L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]\)

\(\displaystyle = \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}\)

\(\displaystyle n^2\) will dominate over \(\displaystyle (n+1)\) since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence. 

Alternatively, it's clear that \(\displaystyle n!\) is much greater than \(\displaystyle n^2\), and thus having \(\displaystyle n!\) in the numerator will make the series diverge by the \(\displaystyle n^{th}\) limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

First we need to set up our system of equations.

\(\displaystyle 2=2\lambda x \rightarrow x=\frac{1}{\lambda}\)

\(\displaystyle -5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}\)

\(\displaystyle x^2+y^2=144\)

Now lets plug in these constraints.

\(\displaystyle (\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144\)

 \(\displaystyle \frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144\)

\(\displaystyle \frac{29}{4\lambda^2}=144\)

Now we solve for \(\displaystyle \lambda\)

\(\displaystyle \frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}\)

If

 \(\displaystyle \lambda=\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46\), \(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14\)

If

 \(\displaystyle \lambda=-\sqrt{\frac{29}{576}}\)

\(\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46\), \(\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14\)

Now lets plug in these values of \(\displaystyle x\), and \(\displaystyle y\) into the original equation.

\(\displaystyle f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62\)

\(\displaystyle f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62\)

We can conclude from this that \(\displaystyle f(4.46,-11.14)\) is a maximum, and \(\displaystyle f(-4.46,11.14)\) is a minimum.

Let \(\displaystyle g = x^2 +2y^2\)To find the absolute minimum value, we must solve the system of equations given by

\(\displaystyle \triangledown f = \lambda\triangledown g, g =1\).

So this system of equations is

\(\displaystyle f_x = \lambda g_x\), \(\displaystyle f_y = \lambda g_y\), \(\displaystyle g =1\).

Taking partial derivatives and substituting as indicated, this becomes

\(\displaystyle 2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1\).

From the left equation, we see either \(\displaystyle x=0\) or \(\displaystyle \lambda = 1\). If \(\displaystyle x=0\), then substituting this into the other equations, we can solve for \(\displaystyle y, \lambda\), and get \(\displaystyle y = \pm \sqrt{2}/2\), \(\displaystyle \lambda = 1/2\), giving two extreme candidate points at \(\displaystyle (0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})\).

On the other hand, if instead \(\displaystyle \lambda =1\), this forces \(\displaystyle y = 0\) from the 2nd equation, and \(\displaystyle x = \pm 1\) from the 3rd equation. This gives us two more extreme candidate points; \(\displaystyle (-1,0),(1,0)\).

Taking all four of our found points, and plugging them back into \(\displaystyle f\), we have

\(\displaystyle f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}\).

Hence the absolute minimum value is \(\displaystyle \frac{1}{2}\).