1. Since both equations contain \(x^2\) and \(y^2\) only, we can use elimination as seen in Section AppLinearSystems:

   \[ \begin{array}{ccc} \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & 4x^2+9y^2 & = & 36 \\ \end{array} \right . & \xrightarrow [\text {$-4E1 + E2$}]{\text {Replace $E2$ with}} & \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & 5y^2 & = & 20 \\ \end{array} \right . \end{array} \]

   From \(5y^2 = 20\), we get \(y^2 = 4\) or \(y = \pm 2\). To find the associated \(x\) values, we substitute each value of \(y\) into one of the equations to find the resulting value of \(x\).

   Choosing \(x^2 + y^2 = 4\), we find that for both \(y=-2\) and \(y=2\), we get \(x=0\). Our solution is thus \(\{(0,2),(0,-2)\}\). To verify these answers algebraically, we would need to show that the pair \((x,y) = (0,2)\) and \((x,y) = (0,-2)\) each satisfy both equations. We leave this to the reader.

   To check our answer graphically, we sketch both equations and look for their points of intersection.

   The graph of \(x^2 + y^2 = 4\) is a circle centered at \((0,0)\) with a radius of \(2\). To graph \(4x^2+9y^2 = 36\), we convert to standard form \(\frac {x^2}{9} + \frac {y^2}{4} = 1\) and recognize it as an ellipse centered at \((0,0)\) with a major axis along the \(x\)-axis of length \(6\) and a minor axis along the \(y\)-axis of length \(4\).

   We see from the graph that the two curves intersect at their \(y\)-intercepts only, \((0, \pm 2)\).

2. We proceed as before to eliminate one of the variables

   \[ \begin{array}{ccc} \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & 4x^2-9y^2 & = & 36 \\ \end{array} \right . & \xrightarrow [\text {$-4E1 + E2$}]{\text {Replace $E2$ with}} & \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & -13y^2 & = & 20 \\ \end{array} \right . \end{array} \]

   Since the equation \(-13y^2=20\) admits no real solution, the system is inconsistent. To verify this graphically, we note that \(x^2+y^2=4\) is the same circle as before, but when writing the second equation in standard form, \(\frac {x^2}{9} - \frac {y^2}{4} = 1\), we find a hyperbola centered at \((0,0)\) opening to the left and right with a transverse axis of length \(6\) and a conjugate axis of length \(4\).

   We see that the circle and the hyperbola have no points in common, hence, there are no solutions.

3. Since there are no like terms among the two equations, elimination won’t work here. Instead, we proceed using substitution.

   From the equation \(y - 2x =0\), we get \(y=2x\). Substituting this into \(x^2+y^2=4\) gives \(x^2+(2x)^2 = 4\). Solving, we find \(5x^2 = 4\) or \(x = \pm \frac {2 \sqrt {5}}{5}\).

   Returning to the equation we used for the substitution, \(y = 2x\), we find \(y = \frac {4 \sqrt {5}}{5}\) when \(x = \frac {2 \sqrt {5}}{5}\), so one solution is \(\left ( \frac {2 \sqrt {5}}{5} , \frac {4 \sqrt {5}}{5} \right )\) and the other is \(\left ( -\frac {2 \sqrt {5}}{5} , -\frac {4 \sqrt {5}}{5} \right )\). Hence, our final answer is \(\left \{\left ( \frac {2 \sqrt {5}}{5} , \frac {4 \sqrt {5}}{5} \right ), \left ( -\frac {2 \sqrt {5}}{5} , -\frac {4 \sqrt {5}}{5} \right ) \right \}\). As before, we leave the algebraic check to the reader.

   The graph of \(x^2+y^2=4\) is our circle from before and the graph of \(y - 2x =0\), or \(y = 2x\) is a line through the origin with slope \(2\). Even though we cannot easily verify the numerical values of the points of intersection from our sketch, we can be sure there are just two solutions: one in Quadrant I and one in Quadrant III. This observation, combined with our (your) algebraic check gives us confidence our solution is correct. ^[1] (Of course, we could check our answers more accurately using a graphing utility.)

4. While it may be tempting to solve \(y-x^2=0\) as \(y=x^2\) and substitute, we note that this system is set up for elimination. ^[2] (We encourage the reader to solve the system using substitution to see that you get the same solution.)

   \[ \begin{array}{ccc} \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & y - x^2 & = & 0 \\ \end{array} \right . & \xrightarrow [\text {$E1 + E2$}]{\text {Replace $E2$ with}} & \left \{\begin{array}{lrcr} (E1) & x^2 +y^2 & = & 4 \\ (E2) & y^2 +y & = & 4 \\ \end{array} \right . \end{array} \]

   From \(y^2 + y = 4\) we get \(y^2+y-4 = 0\) which gives \(y = \frac {-1 \pm \sqrt {17}}{2}\). Due to the complicated nature of these answers, it is worth our time to make a quick sketch of both equations first to head off any extraneous solutions we may encounter.

   We see that the circle \(x^2+y^2=4\) intersects the parabola \(y=x^2\) exactly twice, and both of these points have a positive \(y\) value.

   Of the two solutions for \(y\), only \(y = \frac {-1 + \sqrt {17}}{2}\) is positive, so to get our solution, we substitute this into \(y - x^2 = 0\) and solve for \(x\). We get \(x = \pm \sqrt {\frac {-1 + \sqrt {17}}{2}} = \pm \frac {\sqrt {-2 + 2\sqrt {17}}}{2}\).

   Our final answer is \(\left \{ \left (\frac {\sqrt {-2 + 2\sqrt {17}}}{2},\frac {-1 + \sqrt {17}}{2}\right ),\left (-\frac {\sqrt {-2 + 2\sqrt {17}}}{2},\frac {-1 + \sqrt {17}}{2}\right ) \right \}\). Checking these answers algebraically amounts to a true test of anyone’s algebraic mettle and as such is left to the reader.

1. At first glance, it doesn’t appear as though elimination will do us any good since it’s clear that we cannot completely eliminate one of the variables. The alternative, however, namely solving one of the equations for one variable and substituting it into the other, is very intimidating.

   Returning to elimination, we note that it is possible to eliminate the troublesome \(xy\) term, and the constant term as well. Doing so we get a more tractable relationship between \(x\) and \(y\):

   \[ \begin{array}{ccc} \left \{\begin{array}{lrcr} (E1) & x^2 +2xy -16 & = & 0 \\ (E2) & y^2 +2xy -16 & = & 0 \\ \end{array} \right . & \xrightarrow [\text {$-E1 + E2$}]{\text {Replace $E2$ with}} & \left \{\begin{array}{lrcr} (E1) & x^2 +2xy -16 & = & 0 \\ (E2) & y^2 - x^2 & = & 0 \\ \end{array} \right . \end{array} \]

   We get \(y^2 - x^2 = 0\) or \(y = \pm x\). Substituting \(y=x\) into \(E1\) we get \(x^2+2x^2-16 = 0\) so that \(x^2 = \frac {16}{3}\) or \(x = \pm \frac {4 \sqrt {3}}{3}\). On the other hand, when we substitute \(y = -x\) into \(E1\), we get \(x^2 - 2x^2 - 16 = 0\) or \(x^2 = -16\) which gives no real solutions.

   Substituting each of \(x = \pm \frac {4 \sqrt {3}}{3}\) into \(y=x\) yields the solution \(\left \{\left (\frac {4 \sqrt {3}}{3},\frac {4 \sqrt {3}}{3}\right ), \left (-\frac {4 \sqrt {3}}{3},-\frac {4 \sqrt {3}}{3}\right )\right \}\). As usual, we leave it to the reader to verifying this solution algebraically.

   To verify this equation graphically, we solve \(x^2 +2xy -16 = 0\) for \(y\) to obtain \(y = \frac {16 - x^2}{2x}\). We produce the graph of this equation using the techniques described in Section RationalGraphs.

   Solving the second equation, \(y^2 +2xy -16 = 0\), for \(y\), however, is more complicated. The quadratic formula gives \(y = -x \pm \sqrt {x^2+16}\) which requires Calculus or a graphing utility to graph.

   As it happens, however, we don’t need either because the equation \(y^2+2xy-16 = 0\) can be obtained from the equation \(x^2+2xy-16=0\) by interchanging ‘\(y\)’ and ‘\(x\).’ Thinking back to Section InverseFunctions, this means we can obtain the graph of \(y^2+2xy-16 = 0\) by reflecting the graph of \(x^2+2xy-16 = 0\) across the line \(y=x\). Doing so confirms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required. ^[3] (An amazing coincidence. Of course, we could use a graphing utility to verify our solutions if this didn’t happen to be the case.)

2. Unlike the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving \(y+4e^{2x}=1\) for \(y\), we get \(y = 1 - 4e^{2x}\) which, when substituted into the second equation, yields \(\left (1 - 4e^{2x}\right )^2 + 2e^{x} = 1\).

   After expanding and gathering like terms, we get \(16e^{4x}-8e^{2x} + 2e^{x} = 0\). Factoring gives us \(2e^{x} \left (8e^{3x}-4e^{x} + 1\right ) = 0\), and since \(2e^{x} \neq 0\) for any real \(x\), we are left with solving \(8e^{3x}-4e^{x} + 1=0\).

   We have three terms, and even though this is not a ‘quadratic in disguise’, we can benefit from the substitution \(u = e^{x}\). The equation becomes \(8u^3-4u+1=0\). Using the techniques set forth in Section RealZeros, we find \(u = \frac {1}{2}\) is a zero and factor \(8u^3-4u+1\). We find \(8u^3-4u+1=0\) is equivalent to \(\left (u - \frac {1}{2}\right ) \left (8u^2+4u-2\right ) =0\). So in addition to \(u = \frac {1}{2}\), we need to solve \(8u^2+4u-2=0\).

   We use the quadratic formula to solve \(8u^2+4u-2=0\) and find \(u = \frac {-1 \pm \sqrt {5}}{4}\). Since \(u = e^{x}\), we now must solve \(e^{x} = \frac {1}{2}\) and \(e^{x} = \frac {-1 \pm \sqrt {5}}{4}\). From \(e^{x} = \frac {1}{2}\), we get \(x = \ln \left (\frac {1}{2}\right ) = -\ln (2)\).

   As for \(e^{x} = \frac {-1 \pm \sqrt {5}}{4}\), we first note that \( \frac {-1 - \sqrt {5}}{4} < 0\), so \(e^{x} = \frac {-1 - \sqrt {5}}{4}\) has no real solutions. We are left with \(e^{x} = \frac {-1 + \sqrt {5}}{4}\), so that \(x = \ln \left (\frac {-1 + \sqrt {5}}{4}\right )\).

   We now return to \(y = 1 - 4e^{2x}\) to find the accompanying \(y\) values for each of our solutions for \(x\).

   For \(x=-\ln (2)\):

   \[\begin{array}{rcl} y & = & 1 - 4e^{2x}\\ & = & 1 - 4e^{-2\ln (2)} \\ & = & 1 - 4e^{\ln \left (\frac {1}{4}\right )} \\ & = & 1 - 4\left (\frac {1}{4}\right ) \\ & = & 0 \\ \end{array}\]

   Hence, one solution is \((-\ln (2), 0)\).

   Next, for \(x = \ln \left (\frac {-1 + \sqrt {5}}{4}\right )\):

   \[\begin{array}{rcl} y & = & 1 - 4e^{2x} \\ & = & 1 - 4e^{2\ln \left (\frac {-1 + \sqrt {5}}{4}\right )} \\ & = & 1 - 4e^{\ln \left (\frac {-1 + \sqrt {5}}{4}\right )^2} \\ & = & 1 - 4\left ( \frac {-1 + \sqrt {5}}{4} \right )^2 \\ & = & 1 - 4\left (\frac {3-\sqrt {5}}{8}\right ) \\ & = & \frac {-1+\sqrt {5}}{2}\\ \end{array}\]

   This gives the other solution: \(\left (\ln \left (\frac {-1 + \sqrt {5}}{4}\right ),\frac {-1+\sqrt {5}}{2}\right )\)

   It is a good review of the properties of logarithms to verify both solutions algebrically, so we leave that to the reader.

   While we are able to sketch \(y = 1 - 4e^{2x}\) using the techniques in Section ExponentialFunctions, the second equation is more difficult and so we opt to use Desmos. We see the two graphs intersect at \((-0.6931,0) \approx (-\ln (2), 0)\) and \((-1.1744, 0.618) \approx \left (\ln \left (\frac {-1 + \sqrt {5}}{4}\right ),\frac {-1+\sqrt {5}}{2}\right )\).

3. Our last system involves three variables and provides an opportunity to gain some insight on how to keep such systems organized. Labeling the equations as before, we have

   \[\left \{\begin{array}{lrcl} E1 & z(x-2) & = & x \\ E2 & yz & = & y \\ E3 & (x-2)^2+y^2 & = & 1 \end{array} \right .\]

   The easiest equation to start with appears to be \(E2\). While it may be tempting to divide both sides of \(E2\) by \(y\), we caution against this practice because it presupposes \(y \neq 0\). Instead, we take \(E2\) and rewrite it as \(yz-y = 0\) so \(y(z-1) = 0\). From this, we get two cases: \(y = 0\) or \(z = 1\).

   CASE 1: \(y = 0\). Substituting \(y=0\) into \(E1\) and \(E3\), we get

   \[\left \{\begin{array}{lrcl} E1 & z(x-2) & = & x \\ E3 & (x-2)^2 & = & 1 \end{array} \right .\]

   Solving \(E3\) for \(x\) gives \(x = 1\) or \(x=3\). Substituting these values into \(E1\) gives \(z=-1\) when \(x=1\) and \(z = 3\) when \(x=3\). We obtain two solutions, \((1,0,-1)\) and \((3,0,3)\).

   CASE 2: \(z = 1\). Substituting \(z=1\) into \(E1\) and \(E3\) gives us

   \[\left \{\begin{array}{lrcl} E1 & (1)(x-2) & = & x \\ E3 & (x-2)^2+y^2 & = & 1 \end{array} \right .\]

   Equation \(E1\) gives us \(x-2 = x\) or \(-2 = 0\), which is a contradiction. This means we have no solution to the system in this case, even though \(E3\) is satisfied by infinitely many pairs of points \((x,y)\).

   Hence, our final answer is \(\left \{ (1,0,-1), (3,0,3) \right \}\). These points are easy enough to check algebraically in our three original equations, so that is left to the reader.

   As for verifying these solutions graphically, they require plotting surfaces in three dimensions and looking for intersection points. While this is beyond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points, \((1,0,-1)\) and \((3,0,3)\) courtesy of GeoGebra.

Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles.

We are told Carl can canoe at a rate of \(6\) miles per hour in still water. How does this figure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling \(R\). When traveling downstream, the river is helping Carl along, so we add these two speeds:

So our downstream speed is \((6+R) \frac {\text {miles}}{\text {hour}}\). Substituting this into our ‘distance-rate-time’ equation for the downstream part of the trip, we get:

When traveling upstream, Carl works against the current. Since Carl must move the canoe faster than the river’s speed to move upstream, we subtract the river’s speed from Carl’s canoing speed to get:

Proceeding as before, we get

The last piece of information given to us is that the total trip lasted \(3\) hours. If we let \(t_{\text {down}}\) denote the time of the downstream trip and \(t_{\text {up}}\) the time of the upstream trip, we have: \(t_{\text {down}} + t_{\text {up}} = 3 \, \text {hours}\). Substituting \(t_{\text {down}}\) and \(t_{\text {up}}\) into the ‘distance-rate-time’ equations, we get a system of three equations in three unknowns below. Note that since the variables in equations \(E1\) and \(E2\) are multiplied together, these two equations are nonlinear.

Since we are ultimately after \(R\), we need to use these three equations to get at least one equation involving only \(R\). We start with equation \(E1\). We know that both \((6+R) \neq 0\) and \(t_{\text {down}} \neq 0\) since the product of these two quantities is \(5\) and is nonzero. ^[4] (This is a restatement of the Zero Product Property from Section AppRealNumberArithmetic.) Hence, we may solve \(E1\) for \(t_{\text {down}}\) by dividing both sides by the quantity \((6+R)\) to get \(t_{\text {down}} = \frac {5}{6+R}\). Similarly, we use \(E2\) to get \(t_{\text {up}} = \frac {5}{6-R}\). Substituting these into \(E3\), we get: ^[5] (The reader is encouraged to verify that the units in this equation are consistent. For starters, the units on the ‘3’ is ‘hours.’)

We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then:

So we have that the rate Taylor works is \(\frac {1 \, \text {garden}}{ 4 \, \text {hours}} = \frac {1}{4} \frac {\text {garden}}{\text {hour}}\). We are also told that when working together, Taylor and Carl can weed the garden in just 3 hours. We have:

From this, we find that the rate of Taylor and Carl working together is \(\frac {1 \, \text {garden}}{3 \, \text {hours}} = \frac {1}{3} \frac {\text {garden}}{\text {hour}}\). We are asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown \(t\), measured in hours to be consistent with the other times given to us in the problem. Then:

In order to find \(t\), we need to find the rate of Carl working, so let’s call this quantity \(R\), with units \(\frac {\text {garden}}{\text {hour}}\). Using the fact that rates are additive, we have:

so that \(R = \frac {1}{12} \frac {\text {garden}}{\text {hour}}\). Substituting this into our ‘work-rate-time’ equation for Carl, we get:

Solving \(1 = \frac {1}{12} t\), we get \(t = 12\), so it takes Carl 12 hours to weed the garden on his own. ^[6] (Carl would much rather spend his time writing open-source Mathematics texts than gardening anyway.)