Nature of Solutions of a System of Linear Equations

Nature of Solutions of a System of Linear Equations exercise

• Explain how to find the solution to the system of equations.

  Hints

  A linear equation is often written in the form: $y=mx+b$.

  In this case, $m$ is the slope and $b$ the y-intercept.

  If two linear equations have the same slope, they are parallel.

  • If the y-intercepts differ, the lines will never meet.
  • If the y-intercepts are the same, the lines are the same.

  Solution

  Looking at the menu we can see the prices:

  • Adults have to pay $\$~10$ to enter and kids $\$~8$.
  • The normal sushi costs $20$ cents and the fancy sushi $30$ cents per plate.

  This leads to two linear equations. Let $x$ be the number of sushi plates eaten by Jimmy and Mike.

  The non-adventurous uncle Mike

  $y=10+0.20x$ The adventurous Jimmy

  $y=8+0.3x$

  Identifying those equations leads to a linear equation:

  $\begin{array}{rcl} 10+0.20x & = & ~8+0.30x\\ \color{#669900}{-0.30x} & &\color{#669900}{-0.30x}\\ 10-0.10x & = & ~8\\ \color{#669900}{-10} & & \color{#669900}{-10}\\ -0.10x & = & ~-2\\ \color{#669900}{\div (-0.10)} & & \color{#669900}{\div (-0.10)}\\ x & = & ~20 \end{array}$

  The two lines intersect at $x=20$ and $y=10+0.20\times 20=14$.

  If Mike decides to take the fancy sushi, the resulting equation is a bit different:

  $y=10+0.30x$

  Have a look at the slope of the equations for Jimmy and Mike. They have the same slope: $m=0.30$ but different y-intercepts. The two lines are parallel and never meet.

  If the restaurant offers a special day with the same buffet price for adults and kids, and the same price for all types of sushi plates, the corresponding price would be given by $y=9+0.25x$

  Here we can see that the slope and y-intercept are the same for both adults and kids.

  The corresponding lines are the same, so they have infinitely many points in common.

• Find the graph which corresponds to the system of equations.

  Hints

  Given an equation in this form $y=mx+b$, here are what the parameters stand for:

  • $m$ is the slope
  • $b$ is the y-intercept

  • A positive slope leads to an increasing line.
  • A negative slope leads to a decreasing line.

  Solution

  To detect which line corresponds to a given equation, first look at the y-intercept: This is where the line passes the y-axis.

  Let's start with the red line passing through the origin. Thus $b=0$. The corresponding equation is given by $y=x$.

  Next, we examine the blue line crossing the y-axis at $3.5$. The slope is negative. So the corresponding line must be $y=3.5-0.5x$.

  The green line crosses the y-axis at $1$ and is increasing: The corresponding equation is given by $y=1+0.5x$.

  Last, we consider the orange line crossing the y-axis at $-0.5$. The line is increasing: $y=-0.5+1.5x$.

• Determine which systems of equations have one, infinitely many, or no solutions.

  Hints

  Two equations with different slopes, $m$, have only one point in common.

  Two equations with the same slope, represented by $m$, have either no points or infinitely many points in common.

  This depends on the y-intercept.

  Take a look at this example:

  • The slope $m=2$ is the same.
  • The y-intercepts are different. So the lines have no points in common.

  Solution

  The type of solutions of a system of linear equations depends on the slope and the y-intercept:

  • If both are the same, the lines have infinitely many points in common.
  • If just the slopes are the same, the lines are parallel and thus have no points in common.
  • If the slopes are different, there is just one point in common.

  One solution

  • $y=2x+3$; $y=3x+2$
  • $y=0.5x$; $y=-0.5x+2$

  No solution

  • $y=2x+3$; $y=2x+2$
  • $y=0.5x$; $y=0.5x+2$

  Infinitely many solutions

  • $y=2x+3$; $y=2x+3$
  • $y=0.5x$; $y=0.5x$

• Graph the system of equations and find the solution(s), if any exist.

  Hints

  To draw a line corresponding to a given linear equation, first draw the the y-intercept in a coordinate plane.

  Starting at the y-intercept, you can use rise over run to draw the slope of the line.

  The given lines have one point in common.

  You can determine this point algebraically by solving the equation $12+0.40x=9+0.60x$.

  Just plug the given points in the equations to check if they are on the correct lines.

  Solution

  The line $y=12+0.40x$ is the blue one and the line $y=9+0.60x$ is the red one.

  You can recognize those lines by first looking at the y-intercept and checking the given points after.

  The intersection of those lines, and thus the solution of the system of equations, is $(15,18)$.

  You can determine this solution algebraically:

  $\begin{array}{rcl} 12+0.40x & = & ~9+0.60x\\ \color{#669900}{-0.60x} & &\color{#669900}{-0.60x}\\ 12-0.20x & = & ~9\\ \color{#669900}{-12} & & \color{#669900}{-12}\\ -0.20x & = & ~-3\\ \color{#669900}{\div (-0.20)} & & \color{#669900}{\div (-0.20)}\\ x & = & ~15 \end{array}$

  Finally, you get the y-coordinate of the solution by plugging this value in one of the given equations. It doesn't matter which one you choose:

  $y=12+0.40\times 15=12+6=18$ $~~~~~$✓

• Identify which statements are true.

  Hints

  Here you see two lines with different slopes and different y-intercepts.

  Any two given lines have either one point, no points or infinitely many points in common.

  Solution

  The type of solutions to systems of linear equations depends on the slope as well as the y-intercept of the equations:

  Let's have two equations

  $y=m_1x+b_1$ and

  $y=m_2x+b_2$

  First, let's have a look at the slope of the lines:

  • $m_1\neq m_2$ $\rightarrow$ the lines have one point in common independently on the y-intercept.
  • $m_1=m_2$

  Here we can differentiate between two cases:

  • $b_1=b_2$ $\rightarrow$ the lines are identical and thus have infinitely many points in common.
  • $b_1\neq b_2$ $\rightarrow$ the lines are parallel and never meet. They have no points in common.

• Find the solution(s), if any exist, to the system of equations.

  Hints

  If necessary, transform the equation into the form:

  $y=mx+b$.

  • If only the slope of two equations is the same, the lines are parallel and never meet. So you get no solution.
  • If both the slope and the y-intercept are the same, there are infinitely many solutions.

  If two equations have different slopes you get the unique solution by setting the $y$ variables equal to each other and solving for $x$.

  Solution

  To check which type of solution two (or more) equations have, we just examine the slope and the y-intercept:

  • If both are the same, there are infinitely many solutions.
  • If just the slope is the same, there exist no solutions.

  If the equations have different slopes we get the (unique!) solution algebraically.

  Form left to the right:

  The farthest to the left: The slope $m=2$ is the same but the y-intercepts differ. Thus this system has no solution.

  The second equations from the left: The first and the second equation are the same. The third one has a different slope, so you get the solution solving the following equation:

  $\begin{array}{rcl} 0.5x+10 & = & ~-0.5x-10\\ \color{#669900}{+0.5x} & &\color{#669900}{+0.5x}\\ x+10 & = & ~-10\\ \color{#669900}{-10} & & \color{#669900}{-10}\\ x & = & ~-20 \end{array}$

  This leads to $y=0.5\times (-20)+10=-10+10=0$

  The equations in the middle: This system of equations also has a unique solution.

  $\begin{array}{rcl} -2x-5 & = & ~2x+3\\ \color{#669900}{-2x} & &\color{#669900}{-2x}\\ -4x-5 & = & ~3\\ \color{#669900}{+5} & & \color{#669900}{+5}\\ -4x & = & ~8\\ \color{#669900}{\div(-4)} & & \color{#669900}{\div(-4)}\\ x & = & ~-2 \end{array}$

  Again, we get $y$ by plugging in the value of $x$ we found into either of the two equations:

  $\quad~~~y=2\times (-2)+3=-4+3=-1$

  The second equations from the right: Just look at the equations:

  $\begin{array}{rcl} y&=&~2x+3\\ 2y-4x&=&~-6 \end{array}$

  What's that? The second equation looks a little different from the first one. So we first transform this equation:

  • Add $4x$ to get $2y=4x-6$.
  • Divide by $2$. This leads to $y=2x-3$.

  So, the slope is the same, but the y-intercept is different. Therefore this system of equations has no solutions.

  The equations to the far right: If you multiply the second equation by $-1$ you just get the first one. So both equations are the same. Thus, this system has infinitely many solutions.