Solving Systems Using Substitution Section 6-2

Goals Goal Rubric To solve systems of equations by using substitution. Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

Vocabulary Substitution Method

Substitution Method Sometimes it is difficult to identify the exact solution to a system by graphing. In that case, you can use an algebraic method called substitution to solve a system. Substitution is used to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.

Substitution Method Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. Substitution is used to reduce the system to one equation that has only one variable. Then you can solve this equation for the one variable and substitute again to find the other variable.

Solving Systems of Equations by Substitution Substitution Method Solving Systems of Equations by Substitution Step 2 Step 3 Step 4 Step 5 Solve for one variable in at least one equation, if necessary. Step 1 Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute that value into one of the original equations and solve for the other variable. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

Example: Substitution Solve the system by substitution. y = 3x y = x – 2 Step 1 y = 3x Both equations are solved for y. y = x – 2 Step 2 y = x – 2 3x = x – 2 Substitute 3x for y in the second equation. Step 3 –x –x 2x = –2 2x = –2 2 2 x = –1 Now solve this equation for x. Subtract x from both sides and then divide by 2.

Example: Continued   Solve the system by substitution. Step 4 y = 3x Write one of the original equations. Step 4 y = 3x y = 3(–1) y = –3 Substitute –1 for x. Write the solution as an ordered pair. Step 5 (–1, –3) Check Substitute (–1, –3) into both equations in the system. y = 3x –3 3(–1) –3 –3  y = x – 2 –3 –1 – 2 –3 –3 

You can substitute the value of one variable into either of the original equations to find the value of the other variable. Helpful Hint

Example: Substitution Solve the system by substitution. y = x + 1 4x + y = 6 The first equation is solved for y. Step 1 y = x + 1 Write the second equation. Step 2 4x + y = 6 Substitute x + 1 for y in the second equation. 4x + (x + 1) = 6 5x + 1 = 6 Simplify. Solve for x. Step 3 –1 –1 5x = 5 Subtract 1 from both sides. 5x = 5 5 5 Divide both sides by 5. x = 1

Example: Continued Solve the system by substitution. Step 4 y = x + 1 Write one of the original equations. Step 4 y = x + 1 y = 1 + 1 y = 2 Substitute 1 for x. Step 5 (1, 2) Write the solution as an ordered pair.

Example: Substitution Solve the system by substitution. x + 2y = –1 x – y = 5 Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. −2y −2y x = –2y – 1 Step 2 x – y = 5 (–2y – 1) – y = 5 Substitute –2y – 1 for x in the second equation. –3y – 1 = 5 Simplify.

Example: Continued Step 3 –3y – 1 = 5 +1 +1 –3y = 6 –3y = 6 –3 –3 Solve for y. +1 +1 –3y = 6 Add 1 to both sides. –3y = 6 –3 –3 y = –2 Divide both sides by –3. Step 4 x – y = 5 Write one of the original equations. x – (–2) = 5 x + 2 = 5 Substitute –2 for y. –2 –2 x = 3 Subtract 2 from both sides. Write the solution as an ordered pair. Step 5 (3, –2)

Your Turn: Solve the system by substitution. Check your answer. y = x + 3 y = 2x + 5 Step 1 y = x + 3 y = 2x + 5 Both equations are solved for y. Step 2 2x + 5 = x + 3 y = x + 3 Substitute 2x + 5 for y in the first equation. –x – 5 –x – 5 x = –2 Step 3 2x + 5 = x + 3 Solve for x. Subtract x and 5 from both sides.

Continued Solve the system by substitution. Check your answer. Step 4 Write one of the original equations. Step 4 y = x + 3 y = –2 + 3 y = 1 Substitute –2 for x. Step 5 (–2, 1) Write the solution as an ordered pair. Check Substitute (–2, 1) into both equations in the system. y = x + 3 1 (–2) + 3 1 1  y = 2x + 5 1 2(–2) + 5 1 –4 + 5  1 1

Your Turn: Solve the system by substitution. x = 2y – 4 Step 1 x = 2y – 4 The first equation is solved for x. (2y – 4) + 8y = 16 x + 8y = 16 Step 2 Substitute 2y – 4 for x in the second equation. Step 3 10y – 4 = 16 Simplify. Then solve for y. +4 +4 10y = 20 Add 4 to both sides. 10y 20 10 10 = Divide both sides by 10. y = 2

Continued x + 8y = 16 Step 4 x + 16 = 16 x = 0 Step 5 (0, 2) Write one of the original equations. x + 8(2) = 16 Substitute 2 for y. x + 16 = 16 Simplify. x = 0 – 16 –16 Subtract 16 from both sides. Write the solution as an ordered pair. Step 5 (0, 2)

Your Turn: Solve the system by substitution. 2x + y = –4 x + y = –7 Step 1 x + y = –7 – y – y x = –y – 7 Solve the second equation for x by subtracting y from each side. 2(–y – 7) + y = –4 x = –y – 7 Step 2 Substitute –y – 7 for x in the first equation. 2(–y – 7) + y = –4 Distribute 2. –2y – 14 + y = –4

Continued Step 3 –2y – 14 + y = –4 –y – 14 = –4 +14 +14 –y = 10 Combine like terms. –y – 14 = –4 +14 +14 –y = 10 Add 14 to each side. y = –10 Step 4 x + y = –7 Write one of the original equations. x + (–10) = –7 Substitute –10 for y. x – 10 = – 7

Continued Step 5 x – 10 = –7 +10 +10 x = 3 Step 6 (3, –10) +10 +10 Add 10 to both sides. x = 3 Step 6 (3, –10) Write the solution as an ordered pair.

When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. You have to use both equations. Caution

Example: Substitution Involving Distribution y + 6x = 11 Solve by substitution. 3x + 2y = –5 Solve the first equation for y by subtracting 6x from each side. Step 1 y + 6x = 11 – 6x – 6x y = –6x + 11 3x + 2(–6x + 11) = –5 3x + 2y = –5 Step 2 Substitute –6x + 11 for y in the second equation. 3x + 2(–6x + 11) = –5 Distribute 2 to the expression in parentheses.

Continued 3x + 2(–6x + 11) = –5 Step 3 3x – 12x + 22 = –5 Simplify. Solve for x. 3x – 12x + 22 = –5 –9x + 22 = –5 –9x = –27 – 22 –22 Subtract 22 from both sides. –9x = –27 –9 –9 Divide both sides by –9. x = 3

Continued x = 3 Step 4 y + 6x = 11 y + 6(3) = 11 y + 18 = 11 –18 –18 Write one of the original equations. Step 4 y + 6x = 11 y + 6(3) = 11 Substitute 3 for x. y + 18 = 11 Simplify. –18 –18 y = –7 Subtract 18 from each side. Step 5 (3, –7) Write the solution as an ordered pair.

Your Turn: –2x + y = 8 Solve by substitution. 3x + 2y = 9 Step 1 –2x + y = 8 Solve the first equation for y by adding 2x to each side. + 2x +2x y = 2x + 8 3x + 2(2x + 8) = 9 3x + 2y = 9 Step 2 Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9 Distribute 2 to the expression in parentheses.

Continued Step 3 3x + 4x + 16 = 9 7x + 16 = 9 –16 –16 7x = –7 7x = –7 Simplify. Solve for x. 3x + 4x + 16 = 9 7x + 16 = 9 7x = –7 –16 –16 Subtract 16 from both sides. 7x = –7 7 7 Divide both sides by 7. x = –1

Continued x = –1 Step 4 –2x + y = 8 –2(–1) + y = 8 y + 2 = 8 –2 –2 Write one of the original equations. Step 4 –2x + y = 8 –2(–1) + y = 8 Substitute –1 for x. y + 2 = 8 Simplify. –2 –2 y = 6 Subtract 2 from each side. Step 5 (–1, 6) Write the solution as an ordered pair.

Example: A Linear System with No Solution Show that this linear system has no solution. 2 x  y  5 Equation 1 2 x  y  1 Equation 2 METHOD: SUBSTITUTION Because Equation 2 can be revised to y  –2 x  1, you can substitute –2 x  1 for y in Equation 1. 2 x  y  5 Write Equation 1. 2 x  (–2 x  1)  5 Substitute –2 x  1 for y. 1  5 Simplify. False statement! Once the variables are eliminated, the statement is not true regardless of the values of x and y. This tells you the system has no solution.

Example: A Linear System with Many Solutions Show that this linear system has infinitely many solutions. – 2 x  y  3 Equation 1 – 4 x  2y  6 Equation 2 METHOD: SUBSTITUTION You can solve Equation 1 for y. y  2 x + 3 Substitute 2 x + 3 for y in Equation 2. – 4 x  2( 2 x + 3)  6 – 4 x + 4x  6  6 Simplify. 6  6 True statement! The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This result tells you that the linear system has infinitely many solutions.

Your Turn: Solve the system. 30 = 30 (true statement) Infinite Solutions 6 = -9 (false statement) No Solution

Joke Time Why was the sand wet? Because the sea weed! Why did the man dump ground beef on his head? He wanted a meatier shower! What do eskimos get from sitting on the ice too long? Polaroids.

Assignment 6-2 Exercises Pg. 394 - 395: #12 – 42 even