Introduction to Systems of Equations (and Solving by Graphing) Unit 5 Day 1

Systems Two or more linear equations together form a system of linear equations. Example: A solution of the system of linear equations is any ordered pair that makes all the equations true. There are several ways to find the solution for a system of linear equations. Today, we will explore how to solve by graphing.

Example One (-1, -2) none (1, 3) Looking at these graphs of a system of equations, where do you think the solution is? Solution: _______ Solution: _______ Solution: _______ (-1, -2) none (1, 3)

There are 3 types of solutions ONE SOLUTION (independent system) When the lines intersects at one point. (x, y)

There are 3 types of solutions NO SOLUTION (inconsistent system) When the lines are parallel and never intersect

There are 3 types of solutions INFINITELY MANY SOLUTIONS (dependent system) When the two equations graph the same exact line.

Example Two Without graphing, decide whether the system has one solution, no solution, or infinitely many solutions. a.) y = ½x – 4 b.) y = 3x + 6 c.) y = 2x + 5 y = ½x + 6 -2x + y = – 7 -6x + 3y = 15 y = 2x + 5 y = 2x – 7 Same slope with different y-ints  parallel lines, NO SOLUTION Same slopes, Same y-ints, same line! INFINITELY MANY SOLUTIONS Different slopes, different y-ints, ONE SOLUTION

Example Three: Teacher y = -1/2x + 2 3x + y = -3 (subtract 3x to solve for y) y = -3 – 3x Solution: ( -2 , 3 )

Example Three: Together y = 2x + 1 -12x + 6y = 6 (add 12x to both sides) 6y = 6 + 12x (divide by 6 to both sides) y = 1 + 2x SAME LINE! Infinitely many solutions

Example Three: Your turn! y = x + 5 y = -4x Let’s check your answer by calculator: “Y=” to put in your equations Press “GRAPH” (ZoomOut if necessary) CALC menu (2nd “TRACE”) Option 5: intersect ENTER : 3 times SOLUTION: (-1, 4)

Example Four Suppose you are testing two fertilizers on bamboo plants A and B, which are growing under identical conditions. Plant A is 6 cm tall and growing at a rate of 4 cm/day. Plant B is 10 cm tall and growing at a rate of 2 cm/day. After how many days will the bamboo plants be the same height? What will their height be? Suppose you are testing two fertilizers on bamboo plants A and B, which are growing under identical conditions. Plant A is 6 cm tall and growing at a rate of 4 cm/day. Plant B is 10 cm tall and growing at a rate of 2 cm/day. After how many days will the bamboo plants be the same height? What will their height be? Question is asking in days and height: x = # days, y = height (in cm) Plant A height = starting at 6 then adding 4 per day y = 6 + 4 x Plant B height = starting at 10 then adding 2 per day y = 10 + 2 x Solve by graphing in your calculator (2, 14)  In 2 days, both plants will be 14 cm.

Example Five Your total $ = starting at $20 then adding $5 per week Suppose you have $20 in your bank account. You start saving $5 each week. Your friend has $5 in his account and is saving $10 each week. Assume that neither you nor your friend makes any withdrawals. After how many weeks will you and your friend have the same amount of money in your accounts? Suppose you have $20 in your bank account. You start saving $5 each week. Your friend has $5 in his account and is saving $10 each week. Assume that neither you nor your friend makes any withdrawals. After how many weeks will you and your friend have the same amount of money in your accounts? Question is asking in weeks and $: x = # weeks, y = total $ saved Your total $ = starting at $20 then adding $5 per week y = 20 + 5 x Your friend’s total $=starting at $5 then adding $10 per week y = 5 + 10 x Solve by graphing in your calculator (be sure to zoom out) (3, 35)  In 3 weeks, both you and your friend will have $35.

Your turn! Non-members cost = $4 per class y = 4 x Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay a $10 fee plus an additional $2 per class. After how many classes will the cost be the same? What is the cost? Question is asking in weeks and cost: x = # classes, y = cost Non-members cost = $4 per class y = 4 x Members cost = fee at $10 then adding $2 per class y = 10 + 2 x Solve by graphing in your calculator (5, 20)  In 5 classes, both members and non-members cost will be $20.