Solving Systems of Equations: Algebra vs. Graphs

Welcome to our exploration of systems of equations!A system of equations is a powerful mathematical tool that helps us solve complex problems using multiple equations together.Let's look at an example system with two equations. Notice how both equations use the same variables, x and y.What makes these equations a system is that they share the same variables and must be solved simultaneously.Systems can vary in complexity, from simple two-equation systems to more complex ones with multiple equations.Systems of equations have numerous real-world applications, from economics to physics and chemistry.Now that we understand what systems of equations are, we're ready to learn different methods for solving them.Let's move on to our first solving method.We'll solve this system using two algebraic methods: substitution and elimination.First, let's use the substitution method. We'll take the expression for y from equation one.We substitute this expression into equation two, replacing y.Now we can solve this equation by combining like terms.This gives us x equals one.Now let's solve the same system using elimination. We'll subtract equation two from equation one.When we subtract these equations, the y terms cancel out.This leaves us with three x equals three, so x equals one, the same result we got before.Now that we know x equals one, we can substitute back into either equation to find y.Using equation one, when x is one, y equals two times one plus one, which is three.To solve this system graphically, we'll plot both equations on the same coordinate plane.Let's start with our first equation: y equals two x plus one.This line has a positive slope of two, meaning it rises two units for every one unit to the right.Now let's add our second equation: y equals negative x plus four.This line has a negative slope of negative one, meaning it falls one unit for every one unit to the right.The solution to our system is the point where these lines intersect. At this point, both equations are satisfied simultaneously.We can verify this solution by plugging the point (1,3) back into both equations.For the first equation, when x is 1, y equals 2 times 1 plus 1, which equals 3.For the second equation, when x is 1, y equals negative 1 plus 4, which also equals 3.This confirms that the point (1,3) is indeed the solution to our system of equations.