UNIT 3 REASONING WITH EQUATIONS Lesson 2: Solving Systems of Equations Instruction
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1 Prerequisite Skills This lesson requires the use of the following skills: graphing equations of lines using properties of equality to solve equations Introduction Two equations that are solved together are called systems of equations. The solution to a system of equations is the point or points that make both equations true. Systems of equations can have one solution, no solutions, or an infinite number of solutions. Finding the solution to a system of equations is important to many real-world applications. Key Concepts There are various methods to solving a system of equations. Two methods include the substitution method and the elimination method. Solving Systems of Equations by Substitution This method involves solving one of the equations for one of the variables and substituting that into the other equation. Substitution Method 1. Solve one of the equations for one of the variables in terms of the other variable. 2. Substitute, or replace the resulting expression into the other equation. 3. Solve the equation for the second variable. 4. Substitute the found value into either of the original equations to find the value of the other variable. Solutions to systems are written as an ordered pair, (x, y). This is where the lines would cross if graphed. If the resulting solution is a true statement, such as 9 = 9, then the system has an infinite number of solutions. This is where lines would coincide if graphed. U3-71
2 If the result is an untrue statement, such as 4 = 9, then the system has no solutions. This is where lines would be parallel if graphed. Check your answer by substituting the x and y values back into the original equations. If the answer is correct, the equations will result in true statements. Solving Systems of Equations by Elimination Using Addition or Subtraction This method involves adding or subtracting the equations in the system so that one of the variables is eliminated. Properties of equality allow us to combine the equations by adding or subtracting the equations to eliminate one of the variables. Elimination Method Using Addition or Subtraction 1. Add the two equations if the coefficients of one of the variables are opposites of each other. 2. Subtract the two equations if the coefficients of one of the variables are the same. 3. Solve the equation for the second variable. 4. Substitute the found value into either of the original equations to find the value of the other variable. Solving Systems of Equations by Elimination Using Multiplication This method is used when one set of variables are neither opposites nor the same. Applying the multiplication property of equality changes one or both equations. Solving a system of equations algebraically will always result in an exact answer. Elimination Method Using Multiplication 1. Multiply each term of the equation by the same number. It may be necessary to multiply the second equation by a different number in order to have one set of variables that are opposites or the same. 2. Add or subtract the two equations to eliminate one of the variables. 3. Solve the equation for the second variable. 4. Substitute the found value into either of the original equations to find the value of the other variable. U3-72
3 Common Errors/Misconceptions finding the value for only one of the variables of the system forgetting to distribute negative signs when substituting expressions for variables forgetting to multiply each term by the same number when solving by elimination using multiplication U3-73
4 Guided Practice Example 1 Solve the following system by substitution. x+ y= 2 x y= 0 1. Solve one of the equations for one of the variables in terms of the other variable. It doesn t matter which equation you choose, nor does it matter which variable you solve for. Let s solve x + y = 2 for the variable y. Isolate y by subtracting x from both sides. x+ y= 2 x x y= 2 x Now we know that y is equal to 2 x. 2. Substitute, or replace 2 x into the other equation, x y = 0. It helps to place parentheses around the expression you are substituting. x y = 0 Second equation of the system x (2 x) = 0 Substitute (2 x) for y. U3-74
5 3. Solve the equation for the second variable. x (2 x) = 0 x 2 + x = 0 Distribute the negative over (2 x). 2x 2 = 0 Simplify. 2x = 2 Add 2 to both sides. x = 1 Divide both sides by Substitute the found value, (x = 1), into either of the original equations to find the value of the other variable. x + y = 2 First equation of the system (1) + y = 2 Substitute 1 for x. 1 + y = 2 Simplify. y = 1 Subtract 1 from both sides. The solution to the system of equations is (1, 1). If graphed, the lines would cross at (1, 1). U3-75
6 Example 2 Solve the following system by elimination. 2x 3y= 11 x+ 3y= Add the two equations if the coefficients of one of the variables are opposites of each other. 3y and 3y are opposites, so the equations can be added. Add downward, combining like terms only. 2x 3y= 11 x + 3y= 11 3x + 0 = 0 Simplify. 3x = 0 2. Solve the equation for the second variable. 3x = 0 x = 0 Divide both sides by Substitute the found value, x = 0, into either of the original equations to find the value of the other variable. 2x 3y = 11 First equation of the system 2(0) 3y = 11 Substitute 0 for x. 3y = 11 Simplify. 11 y = Divide both sides by 3. 3 U3-76
7 4. The solution to the system of equations is 0, If graphed, the lines would cross at 0, Example 3 Solve the following system by multiplication. x 3y= 5 2x+ 6y= 4 1. Multiply each term of the equation by the same number. The variable x has a coefficient of 1 in the first equation and a coefficient of 2 in the second equation. Multiply the first equation by 2. x 3y = 5 Original equation 2(x 3y = 5) Multiply the equation by 2. 2x 6y = Add or subtract the two equations to eliminate one of the variables. 2x 6y= 10 + ( 2x+ 6y= 4) 0+ 0= 14 U3-77
8 3. Simplify =14 0 = 14 This is NOT a true statement. x 3y= 5 4. The system does not have a solution. There are 2x+ 6y= 4 no points that will make both equations true. U3-78
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