16.1: Systems of Linear Equations - Two Variables

Substitute the ordered pair \((5,1)\) into both equations.\[\begin{array}{rrclcl} & (5) + 3(1) & = & 8 & & \\[6pt] \implies & 8 & = & 8 & \quad & \text{True} \\[6pt] & 2(5) - 9 & = & (1) & & \\[6pt] \implies & 1 & = & 1 & \quad & \text{True} \\[6pt] \end{array}\nonumber\]The ordered pair \((5,1)\) satisfies both equations, so it is the solution to the system.

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure \(\PageIndex{3}\).

Solve the first equation for \(y\).\[\begin{array}{rrcl} & 2x + y & = & -8 \\[6pt] \implies & y & = & -2x - 8 \\[6pt] \end{array}\nonumber\]Solve the second equation for \(y\).\[\begin{array}{rrcl} & x - y & = & -1 \\[6pt] \implies & y & = & x + 1 \\[6pt] \end{array}\nonumber\]Graph both equations on the same set of axes as in Figure \(\PageIndex{4}\).

The lines appear to intersect at the point \((-3,-2)\). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.\[\begin{array}{rrclcl} & 2(-3) + (-2) & = & -8 & & \\[6pt] \implies & -8 & = & -8 & \quad & \text{True} \\[6pt] & (-3) - (-2) & = & -1 & & \\[6pt] \implies & -1 & = & -1 & \quad & \text{True} \\[6pt] \end{array}\nonumber\]The solution to the system is the ordered pair \((-3,-2)\), so the system is independent.

We solve the first equation for \(y\).\[\begin{array}{rrcl} & -x + y & = & -5 \\[6pt] \implies & y & = & x - 5 \\[6pt] \end{array}\nonumber\]Now we can substitute the expression \(x-5\) for \(y\) in the second equation.\[\begin{array}{rrclcl} & 2x - 5y & = & 1 & & \\[6pt] \implies & 2x - 5(x - 5) & = & 1 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 2x - 5x + 25 & = & 1 & \quad & \left( \text{distributing} \right) \\[6pt] \implies & -3x + 25 & = & 1 & \quad & \left( \text{combining like terms} \right) \\[6pt] \implies & -3x & = & -24 & \quad & \left( \text{subtracting }25\text{ from both sides} \right) \\[6pt] \implies & x & = & 8 & \quad & \left( \text{dividing both sides by }-3 \right) \\[6pt] \end{array}\nonumber\]Now, we substitute \(x=8\) into the first equation and solve for \(y\).\[\begin{array}{rrclcl} & -x + y & = & -5 & & \\[6pt] \implies & -(8) + y & = & -5 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & y & = & 3 & \quad & \left( \text{adding }8\text{ to both sides} \right) \\[6pt] \end{array}\nonumber\]Our solution is \((8,3)\).

Check the solution by substituting \((8,3)\) into both equations.\[\begin{array}{rrclcl} & -x + y & = & -5 & & \\[6pt] \implies & -(8) + (3) & = & -5 & \quad & \text{True} \\[6pt] & 2x - 5y & = & 1 & & \\[6pt] \implies & 2(8) - 5(3) & = & 1 & \quad & \text{True} \\[6pt] \end{array}\nonumber\]

Both equations are already set equal to a constant. Notice that the coefficient of \(x\) in the second equation, \(-1\), is the opposite of the coefficient of \(x\) in the first equation, \(1\). We can add the two equations to eliminate \(x\) without needing to multiply by a constant.\[\begin{array}{rccccc} & x & + & 2y & = & -1 \\[6pt] + & -x & + & y & = & 3 \\[6pt] \hline & & & 3y & = & 2 \\[6pt] \end{array}\nonumber\]Now that we have eliminated \(x\), we can solve the resulting equation for \(y\).\[3y=2 \implies y = \dfrac{2}{3}\nonumber\]Then, we substitute this value for \(y\) into one of the original equations and solve for \(x\).\[\begin{array}{rrclcl} & -x + y & = & 3 & & \\[6pt] \implies & -x + \left( \dfrac{2}{3} \right) & = & 3 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -x & = & \dfrac{7}{3} & \quad & \left( \text{subtracting }\frac{7}{3} \text{ from both sides} \right) \\[6pt] \implies & x & = & -\dfrac{7}{3} & \quad & \left( \text{multiplying both sides by }-1 \right) \\[6pt] \end{array}\nonumber\]The solution to this system is \(\left(-\frac{7}{3},\frac{2}{3}\right)\).

Check the solution in the first equation.\[\begin{array}{rrclcl} & x + 2y & = & -1 & & \\[6pt] \implies & \left( -\dfrac{7}{3} \right) + 2\left( \dfrac{2}{3} \right) & = & -1 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -7 + 2\left( 2 \right) & = & -3 & \quad & \left( \text{multiplying both sides by }3 \right) \\[6pt] \implies & -7 + 4 & = & -3 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & -3 & = & -3 & \quad & \left( \text{combining like terms} \right) \\[6pt] \end{array}\nonumber\]Thus, the solution works in the first equation. We leave it to the reader to verify it satisfies the second equation.

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure \(\PageIndex{5}\) to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has \(3x\) in it and the second equation has \(x\). So if we multiply the second equation by \(-3\), the \(x\)-terms will add to zero.\[\begin{array}{rccccc} & x & - & 2y & = & 11 \\[6pt] \implies & -3(x & - & 2y) & = & -3(11) \\[6pt] \implies & -3x & + & 6y & = & -33 \\[6pt] \end{array}\nonumber\]Now, let's add them.\[\begin{array}{rccccc} & 3x & + & 5y & = & -11 \\[6pt] + & -3x & + & 6y & = & -33 \\[6pt] \hline & & & 11y & = & -44 \\[6pt] \implies & & & y & = & -4 \\[6pt] \end{array}\nonumber\]Substituting this into the second equation (in the original system), we get the following:\[\begin{array}{rrclcl} & x - 2y & = & 11 & & \\[6pt] \implies & x - 2(-4) & = & 11 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & x + 8 & = & 11 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & x & = & 3 & \quad & \left( \text{subtracting }8\text{ from both sides} \right) \\[6pt] \end{array}\nonumber\]Our solution is the ordered pair \((3,-4)\) (we leave it to the reader to algebraically check the solution). See Figure \(\PageIndex{6}\).

One equation has \(2x\) and the other has \(5x\). The least common multiple is \(10x\) so we will have to multiply both equations by a constant in order to eliminate one variable. Let's eliminate \(x\) by multiplying the first equation by \(-5\) and the second equation by \(2\).\[\begin{array}{rccccc} & 2x & + & 3y & = & -16 \\[6pt] \implies & -5(2x & + & 3y) & = & -5(-16) \\[6pt] \implies & -10x & - & 15y & = & 80 \\[6pt] & 5x & - & 10y & = & 30 \\[6pt] \implies & 2(5x & - & 10y) & = & 2(30) \\[6pt] \implies & 10x & - & 20y & = & 60 \\[6pt] \end{array}\nonumber\]Then, we add the two equations together.\[\begin{array}{rccccc} & -10x & - & 15y & = & 80 \\[6pt] + & 10x & - & 20y & = & 60 \\[6pt] \hline & & - & 35y & = & 140 \\[6pt] \implies & & & y & = & -4 \\[6pt] \end{array}\nonumber\]Substitute \(y=-4\) into the original first equation.\[\begin{array}{rrclcl} & 2x + 3y & = & -16 & & \\[6pt] \implies & 2x + 3(-4) & = & -16 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 2x - 12 & = & -16 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & 2x & = & -4 & \quad & \left( \text{adding }12\text{ to both sides} \right) \\[6pt] \implies & x & = & -2 & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array}\nonumber\]The solution is \((-2,-4)\) (again, we leave it to the reader to check this). See Figure \(\PageIndex{7}\).

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.\[\begin{array}{rrclcl} & \dfrac{x}{3} & + & \dfrac{y}{6} & = & 3 \\[6pt] \implies & 6\left( \dfrac{x}{3} \right. & + & \left. \dfrac{y}{6} \right) & = & 6(3) \\[6pt] \implies & 2x & + & y & = & 18 \\[6pt] & \dfrac{x}{2} & - & \dfrac{y}{4} & = & 1 \\[6pt] \implies & 4\left( \dfrac{x}{2} \right. & - & \left. \dfrac{y}{4} \right) & = & 4(1) \\[6pt] \implies & 2x & - & y & = & 4 \\[6pt] \end{array}\nonumber\]Add the two equations to eliminate the \(y\)-variable and solve the resulting equation.\[\begin{array}{rrclcl} & 2x & + & y & = & 18 \\[6pt] + & 2x & - & y & = & 4 \\[6pt] \hline & 4x & & & = & 22 \\[6pt] \implies & x & & & = & \dfrac{11}{2} \\[6pt] \end{array}\nonumber\]Substitute \(x = \frac{11}{2}\) into the first equation.\[\begin{array}{rrclcl} & 2x + y & = & 18 & & \\[6pt] \implies & 2\left( \dfrac{11}{2} \right) + y & = & 18 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 11 + y & = & 18 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & y & = & 7 & \quad & \left( \text{subtracting }11\text{ from both sides} \right) \\[6pt] \end{array}\nonumber\]The solution is \(\left(\frac{11}{2},7\right)\). We leave it to the reader to check it in the other equation.

We can approach this problem in two ways. Because one equation is already solved for \(x\), the most obvious step is to use substitution.\[\begin{array}{rrclcl} & x + 2y & = & 13 & & \\[6pt] \implies & (9 - 2y) + 2y & = & 13 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 9 & = & 13 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array}\nonumber\]Clearly, this statement is a contradiction because \(9 \neq 13\). Therefore, the system has no solution.

The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.\[\begin{array}{rrclcl} & x & = & 9 - 2y & & \\[6pt] \implies & 2y & = & -x + 9 & \quad & \left( \text{subtracting }x\text{ from and adding }2y\text{ to both sides} \right) \\[6pt] \implies & y & = & -\dfrac{1}{2}x + \dfrac{9}{2} & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array}\nonumber\]We then convert the second equation expressed to slope-intercept form.\[\begin{array}{rrclcl} & x + 2y & = & 13 & & \\[6pt] \implies & 2y & = & -x + 13 & \quad & \left( \text{subtracting }2y\text{ from both sides} \right) \\[6pt] \implies & y & = & -\dfrac{1}{2}x + \dfrac{13}{2} & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array}\nonumber\]Comparing the equations, we see that they have the same slope but different \(y\)-intercepts. Therefore, the lines are parallel and do not intersect.

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines (with different \(y\)-intercepts) will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure \(\PageIndex{8}\).

With the Elimination Method, we want to eliminate one of the variables by adding the equations. In this case, let's focus on eliminating \(x\). If we multiply both sides of the first equation by \(-3\), then we will be able to eliminate the \(x\)-variable.\[\begin{array}{rccccc} & x & + & 3y & = & 2 \\[6pt] \implies & (-3)\left( x \right. & + & \left. 3y \right) & = & (-3)(2) \\[6pt] \implies & -3x & - & 9y & = & -6 \\[6pt] \end{array}\nonumber\]Now add the equations.\[\begin{array}{cccccc} & -3x & - & 9y & = & -6 \\[6pt] & 3x & + & 9y & = & 6 \\[6pt] \hline + & & & 0 & = & 0 \\[6pt] \end{array}\nonumber\]We can see that there will be an infinite number of solutions that satisfy both equations.

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let's look at what happens when we convert the system to slope-intercept form.\[\begin{array}{rrcl} & x + 3y & = & 2 \\[6pt] \implies & 3y & = & -x + 2 \\[6pt] \implies & y & = & -\dfrac{1}{3}x + \dfrac{2}{3} \\[6pt] & 3x + 9y & = & 6 \\[6pt] \implies & 9y & = & -3x + 6 \\[6pt] \implies & y & = & -\dfrac{1}{3}x + \dfrac{2}{3} \\[6pt] \end{array}\nonumber\]See Figure \(\PageIndex{9}\). Notice the results are the same. The general solution to the system is \(\left(x,-\frac{1}{3}x + \frac{2}{3}\right)\).

Write the system of equations using \(y\) to replace function notation.\[\begin{cases} y & = & 0.85x & + & 35000 \\ y & = & 1.55x \\ \end{cases}\nonumber\]Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x\).\[\begin{array}{rrclcl} & 0.85x + 35000 & = & 1.55x & & \\[6pt] \implies & 35000 & = & 0.7x & \quad & \left( \text{subtracting }0.85x\text{ from both sides} \right) \\[6pt] \implies & 50000 & = & x & \quad & \left( \text{dividing both sides by }0.7 \right) \\[6pt] \end{array}\nonumber\]Then, we substitute \(x=50,000\) into either the cost function or the revenue function.\[1.55 (50000) = 77500\nonumber\]The break-even point is \((50,000,77,500)\).

The profit function is found using the formula \(P(x)=R(x)-C(x)\).\[\begin{array}{rcl} P(x) & = & R(x) - C(x) \\[6pt] & = & 1.55x - (0.85x + 35000) \\[6pt] & = & 0.7x - 35000 \\[6pt] \end{array}\nonumber\]The profit function is \(P(x)=0.7x-35,000\).

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure \(\PageIndex{11}\).

We see from the graph in Figure \(\PageIndex{12}\) that the profit function has a negative value until \(x=50,000\), when the graph crosses the \(x\)-axis. Then, the graph emerges into positive \(y\)-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is \(0\). The area to the left of the break-even point represents operating at a loss.

Let \(c\) = the number of children and \(a\) = the number of adults in attendance.

The total number of people is \(2,000\). We can use this to write an equation for the number of people at the circus that day.\[c+a=2,000\nonumber\]The revenue from all children can be found by multiplying $25.00 by the number of children, \(25c\). The revenue from all adults can be found by multiplying $50.00 by the number of adults, \(50a\). The total revenue is $70,000. We can use this to write an equation for the revenue.\[25c+50a=70,000\nonumber\]We now have a system of linear equations in two variables.\[\begin{cases} c & + & a & = & 2,000 \\ 25c & + & 50a & = & 70,000 \\ \end{cases}\nonumber\]In the first equation, the coefficient of both variables is \(1\). We can quickly solve the first equation for either \(c\) or \(a\). We will solve for \(a\).\[c+a=2,000 \implies a=2,000-c.\nonumber\]Substitute the expression \(2,000-c\) in the second equation for \(a\) and solve for \(c\).\[\begin{array}{rrclcl} & 25c + 50a & = & 70,000 & & \\[6pt] \implies & 25c + 50(2,000 - c) & = & 70,000 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 25c + 100,000 - 50c & = & 70,000 & \quad & \left( \text{distributing} \right) \\[6pt] \implies & -25c + 100,000 & = & 70,000 & \quad & \left( \text{combining like terms} \right) \\[6pt] \implies & -25c & = & -30,000 & \quad & \left( \text{subtracting }100,000\text{ from both sides} \right) \\[6pt] \implies & c & = & 1,200 & \quad & \left( \text{dividing both sides by }-25 \right) \\[6pt] \end{array}\nonumber\]Substitute \(c=1,200\) into the first equation to solve for \(a\).\[\begin{array}{rrclcl} & c + a & = & 2,000 & & \\[6pt] \implies & 1,200 + a & = & 2,000 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & a & = & 800 & \quad & \left( \text{subtracting }1,200\text{ from both sides} \right) \\[6pt] \end{array}\nonumber\]We find that 1,200 children and 800 adults bought tickets to the circus that day.