Systems of Equations

Systems of Equations A system of equations is made up of two or more equations. To “solve the system” is to find the ordered pair (x, y) that make all equations true. This is where the graphs of the equations intersect. For a system of linear equations, there is: one solution if the lines intersect; no solution if the lines never intersect (i.e. the lines are parallel); and infinitely many solutions if the equations give the same line. To solve a system graphically Graph the equations and find the intersection point. To solve a system with substitution Example 1: Solve the system of equations: y = 3x x + 2y = -21 Step 1: Do you have one of the variables isolated in one of the equations? In other words, do you have x = _______ or y = ________? Yes, we have “y = 3x”. If not, isolate a variable from an equation. Step 2: Substitute that variable into the other equation. x + 2y = -21 x + 2(3x) = -21 Substitute (or replace) the y with 3x because they equal. Step 3: Simplify and solve for that variable. x + 6x = -21 7x = -21 x = -3 Step 4: Substitute the variable you solved for into either equation to find the other coordinate. We found x, so we need y. Plug x into either equation to find y. It won’t matter because this is where the lines intersect, so both lines go through the same point. BUT! Be smart about choosing an equation. “y = 3x” is easier to solve for y than “x + 2y = -21.” y = 3x y = 3(-3) Substitute x = -3 y = -9 Solve. Step 5: Write your final answer as an ordered pair. Remember that the x coordinate is always listed first, even if you find the y coordinate first. Solution: (-3, -9) Example 2: Solve the system of equations: x + 5y = -3 3x – 2y = 8 Step 1 isn’t always so easy. Here we don’t have one of the equations as “x = _______” or “y = ______”. In this situation, you need to pick an equation to solve, and a variable to isolate. Again, be smart! Pick the variable that will be the easiest to solve for. The x of the first equation has no coefficient, so there will be nothing to divide. Step 1: Isolate a variable from one of the equations. x + 5y = -3 Choose the easiest equation and the simplest variable. x = -5y – 3 Subtract 5y from both sides. Step 2: Substitute into the other equation. 3x – 2y = 8 3(-5y – 3) – 2y = 8 Substitute x = -5y – 3. -15y – 9 – 2y = 8 Distribute. -17y – 9 = 8 Combine like terms. -17y = 17 Add 9 to both sides. y = -1 Divide both sides by -17. Step 3: Find x by plugging the y value into the equation x = -5y – 3. Again, it won’t matter which equation you plug the x value in. You can use either of the original equations, but work smarter, not harder! You’ve already found an equation for x, so don’t do the hard work again. x = -5y – 3 x = -5(-1) – 3 Substitute y = -1 x=5–3 Multiply x=2 Subtract Step 4: Final answer. (2, -1) Solving Algebraically for No Solution or Infinitely Many Solutions When solving a system algebraically, you will have no solution if all variables cancel and you’re left with a statement that is NEVER true. Example: 0 = 4. No matter what x or y are, this can never work. This means the lines are parallel and never intersect. The system 2x + 2y = 8 will have no solution. x + y = -2 You will have infinitely many solutions if all variables cancel and you’re left with a statement that is ALWAYS true. Example: 2=2. No matter what x or y are, this will always work. This means the equations have the same line and will always intersect. The system 6x – 2y = -4 will have infinitely many solutions. y = 3x + 2 Solving Systems Using Elimination Think of “elimination” as “cancel” or “get rid of.” You are trying to “get rid of” one of the variables through addition, subtract, or multiplication with addition. Part I: Addition Use elimination through addition if the coefficients for x or y are additive inverses (e.g.,2 & -2, 5 & -5, etc.). Example 3: Solve the system of equations: 3x – 5y = -16 2x + 5y = 31 5 & -5 are additive inverses, and those are the coefficients of y. So, by adding the equations together, the y terms will eliminate, or cancel. + 3x – 5y = -16 2x + 5y = 31 5x + 0y = 15 5x = 15 x=3 Add the like terms. Simplify. Divide by 5. Now, plug in x = 3 into either equation to solve for y. Again, it doesn’t matter which equation since this is where the lines intersect. 3x – 5y = -16 3(3) – 5y = -16 Substitute x = 3 9 – 5y = -16 Simplify -5y = -25 Subtract 9 from both sides. y=5 Divide both sides by -5. Final answer: (3, 5) Part II: Subtraction Use subtraction when the coefficients of x or y are equal. Think about subtracting as adding the negative. So, multiply one of the equations by -1, and then add. You are still subtracting the equations, but it’s easier to avoid simple mistakes by adding instead of subtracting. Example 4: Solve the system of equations: 5x + 2y = 6 9x + 2y = 22 Both y variables have a coefficient of 2. So, to cancel the y’s, we will use subtraction. First, choose one of the equations to multiply by -1. It does not matter which. (5x + 2y = 6) (-1) I chose the first equation to multiply by -1. 9x + 2y = 22 Rewrite the second equation so you don’t forget about it. -5x – 2y = -6 Distribute the -1 to every piece of the first equation. + 9x + 2y = 22 Rewrite the second equation. 4x + 0y = 16 Add the like terms. 4x = 16 Simplify x=4 Divide each side by 4. Now we have the x-coordinate, so plug it in to either equation to find the y-coordinate. 5x + 2y = 6 I picked the first equation. 5(4) + 2y = 6 Substitute x = 4. 20 + 2y = 6 Simplify. 2y = -14 Subtract 20 from each side. y = -7 Divide each side by 2. Final answer: (4, -7)