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Section 3.2 Solving Systems of Equations

What if we need to solve an equation with more than one variable? For instance, consider the equation

\begin{equation*} x+3y = 0. \end{equation*}

For such an equation, a solution is a pair of numbers, an \(x\)-value and a \(y\)-value, that make the equation true when they are both plugged in at the same time. We see that if \(y=1\text{,}\) then \(x=-3\) makes the equation true. If \(y=2\text{,}\) then the correct value for \(x\) is \(-6\text{.}\) Clearly we could go on like this, obtaining a different value of \(x\) for any given value of \(y\text{.}\) There is no way for us to say that this equation has a single solution; it has infinitely many. Often in problems the number of solutions is narrowed down by having more equations, which we call a system of equations. For instance, we may consider the system

\begin{equation*} \begin{aligned} x+3y \amp = \amp 0\\ 2x-y \amp = \amp 1. \end{aligned} \end{equation*}

A solution to this system of two equations is again a pair of numbers, one for each variable, that make all the equations true at the same time. For instance \(x=-3\) and \(y=1\) is not a solution to this system; this pair makes the first equation true, but not the second. The pair \(x=1\) and \(y=1\) makes the second equation true, but not the first, so this is also not a solution to the system.**

So how do we find solutions to systems? The most flexible way of solving systems of equations is known as the method of substitution.

Let's see this at work with the system given above. Using the first equation, we can solve for \(x\) in terms of \(y\) as follows:

\begin{align*} x+3y \amp = \amp 0\\ \amp \updownarrow \amp \\ x \amp = \amp -3y. \end{align*}

Now we substitute \(-3y\) every time we see an \(x\) in the second equation and solve for \(y\text{:}\)

\begin{align*} 2(-3y) - y \amp = \amp 1\\ \amp \updownarrow \amp \\ -7y \amp = \amp 1 \\ \amp \updownarrow \amp \\ y \amp = \amp -\frac{1}{7}. \end{align*}

Now we have \(y=-\frac{1}{7}\text{,}\) hence \(x = -3y = \frac{3}{7}\text{.}\) We can now check to make sure we obtained a correct solution by plugging these values into the original system:

\begin{align*} \frac{3}{7} +3\left(\frac{-1}{7}\right) \amp = \amp \frac{3}{7}-\frac{3}{7} = 0 \checkmark \\ 2\left(\frac{3}{7}\right) - \frac{-1}{7} \amp = \amp \frac{6}{7}+\frac{1}{7} = 1 \checkmark \end{align*}

Solve the following systems of equations:

  1. \begin{align*} x^2y \amp = \amp 1 \\ 32xy^2 \amp = \amp x^2 \end{align*}

  2. \begin{align*} x^2+y^2 \amp = \amp 25\\ 3x-4y \amp = \amp 0 \end{align*}

  3. \begin{align*} ab^3 \amp = \amp 2\\ ab^{-1} \amp = \amp 8 \end{align*}

An exam has thirty questions. The professor subtracts eight points for each incorrect answer and adds seven points for each correct answer. If a student ends up receiving zero points, how many questions were answered correctly? Hint: Let \(C\) denote the number of correct answers and \(I\) denote the number of incorrect answers and set up a system of equations.

Typically you need as many equations as variables in order to not have infinitely many solutions. Use substitution to solve the following system of three equations and three unknowns:

\begin{align*} 2x+3y +z \amp = \amp 1\\ x \amp = \amp 3y \\ x+y \amp = z. \end{align*}

Sometimes we need to solve a system of equations in order to manipulate an expression into another form. For instance, an important type of manipulation in Calculus and Differential Equations is known as the Partial Fraction Decomposition. In this manipulation, we take a single fraction and split it as a sum of pieces with simpler, but different, denominators. For instance, let's write
\begin{equation*} \frac{4}{(x-1)(x+2)} \end{equation*}
in the form
\begin{equation*} \frac{A}{x-1} + \frac{B}{x+2} \end{equation*}
for some numbers \(A\) and \(B\text{.}\)
Solution.
These two expressions are to be equivalent for all values of \(x\text{,}\) we need to solve for \(A\) and \(B\text{.}\) To do this, set the expressions equal to one another and do some manipulations.
\begin{align*} \frac{A}{x-1} + \frac{B}{x+2} \amp = \amp \frac{4}{(x-1)(x+2)}\\ \amp \updownarrow \amp (\text{combine fractions on left})\\ \frac{A(x+2) + B(x-1)}{(x-1)(x+2)} \amp = \amp \frac{4}{(x-1)(x+2)}\\ \amp \updownarrow \amp (\text{simplify the expression on the left})\\ \frac{(A+B)x + 2A -B}{(x-1)(x+2)} \amp = \amp \frac{4}{(x-1)(x+2)}. \end{align*}
Now we set the numerators equal:
\begin{equation*} (A+B)x + 2A-B = 4. \end{equation*}
Note that like terms are grouped on both sides. Since there are zero \(x\)'s on the right, we must have \(A+B = 0\text{.}\) Since \(A\) and \(B\) are numbers, we must have \(2A-B = 4\text{.}\) Thus we have a system of equations
\begin{align*} A+B \amp = \amp 0\\ 2A-B \amp = \amp 4. \end{align*}
Using the first equation we have \(B = -A\text{.}\) Substituting into the second equation we see \(3A = 4\text{.}\) Hence, \(A = \frac{4}{3}\) and \(B=-\frac{4}{3}\text{.}\)

Check that

\begin{equation*} \frac{4}{3(x-1)} - \frac{4}{3(x+2)} = \frac{4}{(x-1)(x+2)} \end{equation*}
by finding a common denominator and simplifying into a single fraction.

Solve for \(h\) and \(k\) to put the expression

\begin{equation*} x^2+2x+5 \end{equation*}
in the form
\begin{equation*} (x-h)^2+k \end{equation*}
as follows:

  1. Expand out \((x-h)^2 + k\) (with \(h\) and \(k\) unknown) and group like terms.

  2. Match like terms with \(x^2+2x+5\) to set up a system of equations for \(h\) and \(k\text{,}\) then solve that system of equations.