Chapter 10.8, Problem 49E

(a)

To determine

To find: The solution of the system of equations $\{\begin{array}{l}\log x+\log y=\frac{3}{2}\\ 2\log x-\log y=0\end{array}$ .

Answer to Problem 49E

The solution of the system of equations $\{\begin{array}{l}\log x+\log y=\frac{3}{2}\\ 2\log x-\log y=0\end{array}$ is $\left(3.16,10\right)$ .

Explanation of Solution

Given information:

The system of equation is $\{\begin{array}{l}\log x+\log y=\frac{3}{2}\\ 2\log x-\log y=0\end{array}$ .

Calculation:

Add both the equations in system of equations.

  $\begin{array}{c}\log x+\log y+\left(2\log x-\log y\right)=\frac{3}{2}+0\\ 3\log x=\frac{3}{2}\\ \log x=\frac{1}{2}\\ x={10}^{\frac{1}{2}}\end{array}$

Substitute $\frac{1}{2}$ for $\log x$ in first equation.

  $\begin{array}{l}\log x+\log y=\frac{3}{2}\\ \frac{1}{2}+\log y=\frac{3}{2}\\ \log y=1\\ y={10}^1\end{array}$

The value of $x$ is equal to ${10}^{\frac{1}{2}}=3.16$ and the value of $y$ is ${10}^1=10$ .

Therefore, the solution of the system of equations $\{\begin{array}{l}\log x+\log y=\frac{3}{2}\\ 2\log x-\log y=0\end{array}$ is $\left(3.16,10\right)$ .

(b)

To determine

To find: The solution of the system of equations $\{\begin{array}{c}{2}^x+2^y=10\\ 4^x+4^y=68\end{array}$ .

Answer to Problem 49E

The solution of the system of equations $\{\begin{array}{c}{2}^x+2^y=10\\ 4^x+4^y=68\end{array}$ are $\left(1,3\right)$ and $\left(3,1\right)$ .

Explanation of Solution

Given information:

The system of equation is $\{\begin{array}{c}{2}^x+2^y=10\\ 4^x+4^y=68\end{array}$ .

Calculation:

Simplify the second equation.

  $\begin{array}{c}{4}^x+4^y=68\\ {\left(2^2\right)}^x+{\left(2^2\right)}^y=68\\ {\left(2^x\right)}^2+{\left(2^y\right)}^2=68\\ {\left(2^x\right)}^2+{\left(2^y\right)}^2+2\cdot 2^x\cdot 2^y-2\cdot 2^x\cdot 2^y=68\end{array}$

Further simplify,

  $\begin{array}{c}{\left(2^x\right)}^2+{\left(2^y\right)}^2+2\cdot 2^x\cdot 2^y-2\cdot 2^x\cdot 2^y=68\\ {\left(2^x+2^y\right)}^2-2\cdot 2^x\cdot 2^y=68\end{array}$

Substitute $10$ for $2^x+2^y$ in the above equation.

  $\begin{array}{c}{\left(2^x+2^y\right)}^2-2\cdot 2^x\cdot 2^y=68\\ {\left(10\right)}^2-2\cdot 2^x\cdot 2^y=68\\ 100-2\cdot 2^{x+y}=68\\ -2\cdot 2^{x+y}=-32\end{array}$

Further simplify,

  $\begin{array}{c}-2\cdot 2^{x+y}=-32\\ 2^{x+y}=16\\ 2^{x+y}=2^4\\ x+y=4\end{array}$

So, the value of $x=4-y$ .

Substitute $4-y$ for $x$ in the first equation of system.

  $\begin{array}{c}{2}^x+2^y=10\\ 2^{4-y}+2^y=10\\ 2^4\cdot 2^{-y}+2^y=10\\ \frac{16}{2^y}+2^y=10\end{array}$

Further simplify,

  $\begin{array}{c}\frac{16}{2^y}+2^y=10\\ \frac{16+{\left(2^y\right)}^2}{2^y}=10\\ 16+{\left(2^y\right)}^2=10\left(2^y\right)\\ {\left(2^y\right)}^2-10\left(2^y\right)+16=0\end{array}$

Further simplify,

  $\begin{array}{c}{\left(2^y\right)}^2-10\left(2^y\right)+16=0\\ {\left(2^y\right)}^2-8\left(2^y\right)-2\left(2^y\right)+16=0\\ \left(2^y\right)\left(2^y-8\right)-2\left(2^y-8\right)=0\\ \left(2^y-2\right)\left(2^y-8\right)=0\end{array}$

Substitute $2^y-2=0$ and simplify.

  $\begin{array}{c}{2}^y-2=0\\ 2^y=2^1\\ y=1\end{array}$

Now substitute $2^y-8=0$ and simplify.

  $\begin{array}{c}{2}^y-8=0\\ 2^y=2^3\\ y=3\end{array}$

So, the values of $y$ for the system of equations are $1$ and $3$ . The corresponding values of $x$ for the system of equations are $4-1=3$ and $4-3=1$ .

Therefore, the solution of the system of equations $\{\begin{array}{c}{2}^x+2^y=10\\ 4^x+4^y=68\end{array}$ are $\left(1,3\right)$ and $\left(3,1\right)$ .

(c)

To determine

To find: The solution of the system of equations $\{\begin{array}{c}x-y=3\\ x^3-y^3=387\end{array}$ .

Answer to Problem 49E

The solution of the system of equations $\{\begin{array}{c}x-y=3\\ x^3-y^3=387\end{array}$ are $\left(-5,-8\right)$ and $\left(8,5\right)$ .

Explanation of Solution

Given information:

The system of equation is $\{\begin{array}{c}x-y=3\\ x^3-y^3=387\end{array}$ .

Calculation:

Simplify the second equation.

  $\begin{array}{c}{x}^3-y^3=387\\ \left(x-y\right)\left(x^2+xy+y^2\right)=387\end{array}$

Substitute $3$ for $x-y$ in the above equation.

  $\begin{array}{c}\left(x-y\right)\left(x^2+xy+y^2\right)=387\\ 3\left(x^2+xy+y^2\right)=387\\ x^2+xy+y^2=129\\ x^2+y^2-2xy+xy+2xy=129\end{array}$

Further simplify,

  $\begin{array}{c}{x}^2+y^2-2xy+xy+2xy=129\\ {\left(x-y\right)}^2+3xy=129\end{array}$

Substitute $3$ for $x-y$ in the above equation.

  $\begin{array}{c}{\left(x-y\right)}^2+3xy=129\\ {\left(3\right)}^2+3xy=129\\ 9+3xy=129\\ 3xy=120\end{array}$

Further simplify,

  $\begin{array}{c}3xy=120\\ xy=\frac{120}{3}\\ xy=40\\ y=\frac{40}{x}\end{array}$

So, the value of $y=\frac{40}{x}$ .

Substitute $\frac{40}{x}$ for $y$ in the first equation of system.

  $\begin{array}{c}x-y=3\\ x-\frac{40}{x}=3\\ \frac{x^2-40}{x}=3\\ x^2-40=3x\end{array}$

Further simplify,

  $\begin{array}{c}{x}^2-40=3x\\ x^2-3x-40=0\\ x^2-8x+5x-40=0\\ x\left(x-8\right)+5\left(x-8\right)=0\end{array}$

Further simplify,

  $\begin{array}{c}x\left(x-8\right)+5\left(x-8\right)=0\\ \left(x+5\right)\left(x-8\right)=0\end{array}$

Substitute $x+5=0$ and simplify.

  $\begin{array}{c}x+5=0\\ x=-5\end{array}$

Now substitute $x-8=0$ and simplify.

  $\begin{array}{c}x-8=0\\ x=8\end{array}$

So, the values of $x$ for the system of equations are $-5$ and

  $8$ . The corresponding values of $y$ for the system of equations are $\frac{40}{-5}=-8$ and $\frac{40}{8}=5$ .

Therefore, the solution of the system of equations $\{\begin{array}{c}x-y=3\\ x^3-y^3=387\end{array}$ are $\left(-5,-8\right)$ and $\left(8,5\right)$ .

(d)

To determine

To find: The solution of the system of equations $\{\begin{array}{c}{x}^2+xy=1\\ xy+y^2=3\end{array}$ .

Answer to Problem 49E

The solution of the system of equations $\{\begin{array}{c}{x}^2+xy=1\\ xy+y^2=3\end{array}$ are $\left(1,3\right)$ and $\left(3,1\right)$ .

Explanation of Solution

Given information:

The system of equation is $\{\begin{array}{c}{x}^2+xy=1\\ xy+y^2=3\end{array}$ .

Calculation:

Add both the equations of the system of equations.

  $\begin{array}{c}{x}^2+xy+\left(xy+y^2\right)=1+3\\ x^2+2xy+y^2=4\\ {\left(x+y\right)}^2=4\\ x+y=\sqrt{4}\end{array}$

Further simplify,

  $\begin{array}{c}x+y=\sqrt{4}\\ x+y=\pm 2\end{array}$

Simplify the first equation of system.

  $\begin{array}{c}{x}^2+xy=1\\ x\left(x+y\right)=1\end{array}$

Substitute $\pm 2$ for $x+y$ in the above equation.

  $\begin{array}{c}x\left(x+y\right)=1\\ x\left(\pm 2\right)=1\\ x=\pm \frac{1}{2}\end{array}$

Simplify the second equation of system.

  $\begin{array}{c}xy+y^2=3\\ y\left(x+y\right)=3\end{array}$

Substitute $\pm 2$ for $x+y$ in the above equation.

  $\begin{array}{c}y\left(x+y\right)=3\\ y\left(\pm 2\right)=3\\ y=\pm \frac{3}{2}\end{array}$

Therefore, the solution of the system of equations $\{\begin{array}{c}{x}^2+xy=1\\ xy+y^2=3\end{array}$ are $\left(\frac{1}{2},\frac{3}{2}\right)$ and $\left(-\frac{1}{2},-\frac{3}{2}\right)$ .