![3( 3 / 2 ) + y = 4 3 x 3 + y = 4 1 x 2 9 / 2 + y = 4 - 9 / 2 -4 ½ y = -½ ( 3 / 2 , -½) or (1.5, -0.5) Change 3 to a fraction Multiply the fractions Subtract 9 / 2 [ or 4 ½] from both sides. The solutions in fraction and decimal forms.](https://image.slidesharecdn.com/solving-linear-equations-1221710708988106-9/85/Solving-Linear-Equations-30-320.jpg)
Solving Linear Equations
- 1. Solving Linear Equations Using Graphing Substitution and Elimination Need some homework help? Go to: http://go.hrw.com Keyword: MA1 Homework Help
- 2. What’s the Deal? There are a number of ways to solve groups of linear equations. In this review, we will find points on a coordinate plane that solve linear equations in standard form and y-intercept form.
- 3. Three Parts Part One – Solve linear equations by graphing. Part Two – Solve linear equations by substitution. Part Three – Solve linear equations by elimination.
- 4. Solving for linear equations answers the question: What values of x and y fit into both equations? The answer is usually given in (x,y) format (ie. (-4, 6) or (3,8).
- 5. Remember - Slope intercept form: y = mx + b m = slope b = y-intercept In y = 1/2x – 7 , the place where the line intercepts the y-axis (called the y-intercept) is negative seven (-7). The slope, which is the rise over the run, is ½ (the fraction before the x): Rise = up, or plus one (+1) Run = right, or plus two (+2).
- 6. If the slope is ½ Rise Run = slope = m The rise is 1 and the run is 2. From the origin (0,0), go up 1 and right 2.
- 7. Graphing Systems of equations y = 3x + 1 y = -x + 5 Since both are in y-intercept format (y=mx+b) find the point through which the line intercepts the y-axis. From that point, graph the slope.
- 8. Graph y=3x + 1 In the following slide, you will see +1 graphed as the y-intercept. And the slope rise =3 and run = 1 will be graphed over the y-intercept.
- 9. y = x + 1 y-intercept
- 10. Let’s add y = -x + 5 The slope is -1. Or down one And right one.
- 11. The coordinates of the intersecting point is your solution. The lines inter- cept at (1, 4) so the solution is x=1, y =4. The lines inter- cept at (1, 4) so the solution is x=1, y =4.
- 12. Solve by graphing y = x +3 y = x +1 The next two slides will show the solution.
- 13. The coordinates of the intersecting point is your solution. The lines inter- cept at (-20,-12) so the solution is x= -20, y = -12.
- 14. Now solve equations in standard form. 3 x + 2y = -6 and -3 x + 2y = 6 When graphing, you must convert equations from standard form to y-intercept form. Let’s review that from a previous lesson using the equations above…
- 15. Change 3x + 2y = -6 to y-intercept form 3x + 2y = -6 - 3x -3x 2y = -3x - 6 Now we need to get y isolated. In this case, let’s divide both sides by 2. 2y = -3x - 6 2 2 2 Now simplify. y = - x -3 Subtract -3x from both sides
- 16. Change -3x + 2y = 6 to y-intercept form -3x + 2y = +6 + 3x 3x 2y = 3x + 6 Get y isolated. Divide both sides by 2. 2y = 3x + 6 2 2 2 Now simplify. y = 3 / 2 x + 3 Add 3x to both sides
- 17. Graph the equations: y = - 3 / 2 x -3 and y = 3 / 2 x + 3 x = 2, y = 0 The solution is (2,0)
- 18. End of Part One
- 19. Part Two – Solve linear equations by substitution Need Help? Go online to go.hrw.com See your textbook’s “internet connect” notes
- 20. Here’s what will happen 1) Find the x or y value that is isolated, such as x = y+3. The x-value is isolated. 2) Insert the isolated value into the equation, this allows you to solve for one variable at a time. 3) Find the value of one variable. 4) Insert that value into either equation and solve for the second variable.
- 21. Solve for the following equations. 2x + 8y = 1 x = 2y Step one ( listed on the previous slide ) 1) Find the x or y value that is isolated, such as x = 2y. The x-value is isolated. Since x = 2y, you insert 2y wherever x occurs.
- 22. Step 2) Insert the isolated value into the equation, this allows you to solve for one variable at a time. 2( 2y ) + 8y = 1 4y + 8y = 1 12y=1 12y = 1 12 12 y = 1 / 12 Replace the x with 2y by substitution. Multiply 2*2y. Combine like terms. Divide both sides by 12 Solve for y. Careful! You are only half done. You still have to solve for the other variable!
- 23. Solve for: 2x + 8y = 1 and x = 2y 3) Find the value one variable. This was done on the previous slide. y = 1 / 12 4) Insert that value into either equation and solve for the second variable. x = 2 ( ) or x = (which is in lowest terms) The solution is (1/6, 1/12)
- 24. Try one. 2x + y = 5 x = 7
- 25. x is given as 7. Insert 7 for any occurrence of x to solve for y. First solve for y. 2x + y = 5 2(7) + y = 5 14 + y = 5 -14 -14 y = -9 Since you know that x = 7 and y = -9, insert those values into a coordinate in (x,y) format. Solution: ( 7, -9 )
- 26. Solve for x and y. 3x + y = 4 and 5x – 7y = 11 Notice that y can more easily be isolated in the first equation. The Plan: Let’s isolate y. Then we will use the value for y to substitute for y.
- 27. Subtract 3x from both sides to isolate y. 3x + y = 4 - 3x -3x y = -3x +4 5x – 7y = 11 5x – 7( -3x +4) = 11 Now that we know that y = -3x+4, substitute. Which property gets used next?
- 28. The Distributive Property is used. 5x – 7( -3x +4) = 11 5x +21x -28 = 11 26x – 28 = 11 +28 +28 Distribute -7 to both terms inside the parentheses. -7(-3x) = + 21x -7(4) = -28 Combine like terms. Add 28 to both sides. Next: divide both sides by 26.
- 29. Continuing… 26x = 39 26 = 26 x = 39 / 26 x = 3 / 2 3( 3 / 2 ) + y = 4 Find lowest terms (divide 39 and 26 by the LCM) Using the x value, solve for y in one of the equations.
- 30. 3( 3 / 2 ) + y = 4 3 x 3 + y = 4 1 x 2 9 / 2 + y = 4 - 9 / 2 -4 ½ y = -½ ( 3 / 2 , -½) or (1.5, -0.5) Change 3 to a fraction Multiply the fractions Subtract 9 / 2 [ or 4 ½] from both sides. The solutions in fraction and decimal forms.
- 31. Part Three – Solving equations by Elimination Need some homework help? Go to: http:// go.hrw.com Keyword: MA1 Homework Help
- 32. Solve using substitution after manipulating equations in standard form. 2x + 3y = 21 -3x – 3y = -12 Which value, x or y, should we work with first? This looks like a very long, drawn-out problem. Is there a better way?
- 33. Let’s solve by elimination. This method uses opposites to eliminate one of the variables. Which variable should be eliminated? 2x + 3y = 21 -3x – 3y = -12
- 34. 2x + 3y = 21 -3x – 3y = -12 Notice that the coefficients with the y value are opposites. (+3 and -3). If we combine these two equations together in columns, we can eliminate the y values. We will solve for x and then insert it’s value into one of the original equations to solve for y.
- 35. The steps and explanations 2x + 3y = 21 -3x – 3y = -12 -1x + 0 = 9 -1x + 0= 9 -1 -1 x = -9 Add terms from top to bottom. +2x - 3x +3y - 3y Divide both sides by -1 . Now go back and insert -9 for x.
- 36. 2x + 3y = 21 -3x – 3y = -12 You may pick either one. 2(-9) + 3y = 21 -18 + 3y = 21 (add 18 to both sides) +3y = 39 3 3 y = 13 Solution (-9, 13) -3(-9) – 3y = -12 +27 – 3y = -12 (subtract 27 from both sides) -3y = -39 -3 -3 y = 13
- 37. Try One. -4x + 3y = -1 4x + 6y = 5
- 38. Eliminate the x values. -4x + 3y = -1 4x + 6y = 5 9y = 4 9y = 4 9 9 y = 4 / 9 Solve for x. 4x + 6( 4 / 9 ) = 5 4x + 24 / 9 = 45 / 9 Subtract 21 / 9 from both sides. 4x = 2 1 / 3 Go to the next slide…
- 39. 4x = 21 / 9 Divide both sided by 4. 4x = 21 / 9 4 4 X = X = To divide fractions, multiply by the reciprocal
- 40. Ready to go one more step? What if you don’t have an easy choice. You may find that neither equation has opposite coefficients.
- 41. Let’s try 11x + 2y = -8 and 8x + 3y = 5 Our goal is to eliminate a variable using opposite coefficients. It looks like we should use 2y and 3y since they are smaller numbers. If we multiply both sides of the top equation by -3 and both sides of the bottom by 2, we should get coefficients of 6 and -6.
- 42. Multiply both sides (11x + 2y) = (-8) (8x + 3y) = (5) -3(11x + 2y) = (-8)-3 2(8x + 3y) = (5)2 We’ll put all four values into parentheses. Multiply both sides of the top by -3 Multiply both sides of the second equation by 2.
- 43. Results of the First Steps -3(11x + 2y) = (-8)-3 2(8x + 3y) = (5)2 ----------------------- -33x – 6y = +24 16x + 6y = +10 -17x + 0 = 34 From the previous slide Use the distributive property Now eliminate
- 44. -17x = 34, x = -2 11x + 2y = -8 11(-2) + 2y = -8 -22+ 2y = -8 2y = 14 y = 7 Pick one of the original equations. Solve for the other variable. Add 22 to both sides. -8 +22 = 14. Solution (-2, 7)
- 45. One more for practice 3x - 2y = 2 4x – 7y = 33 -------------------- -4(3x - 2y) = (2)-4 3(4x – 7y) = (33)3 ------------------------- Solution on the next slide…
- 46. One more for practice 3x - 2y = 2 4x – 7y = 33 -------------------- -4(3x - 2y) = (2)-4 3(4x – 7y) = (33)3 ------------------------- -12x + 8y = -8 12x – 21y = 99 ----------------------- -13y = 91 -13y = 91 -13 -13 y= -7 --------------------------- 3x-2(-7)= 2 3x + 14 = 2 3x = -12 x= -4 ------------- Solution (-4, -7)
- 47. Which way of solving works best for you? Graphing? Substitution? Elimination? Make sure you know them all in order to pick the best way to solve each problem.
- 48. You have reviewed the first three parts of Chapter 7.