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Solving Systems of Linear Equations Overview

Learn how to solve systems of linear equations using substitution and elimination methods with practical examples.

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Solving Systems of Linear Equations Overview

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  1. MA 1128: Lecture 09 – 2/24/11 Solving Systems of Linear Equations

  2. Systems of Linear Equations We are often faced with several equations, and we want to find a solution that works for all of them. We’ll look first at the situation where we have a system of two linear equations. For example, suppose we have the two equations. 3x + 2y = 12 and y = 3x – 3. Note that (2,3) satisfies both equations. 3(2) + 2(3) = 6 + 6 = 12, so it’s a solution to the first equation, and 3(2) – 3 = 6 – 3 = 3, so it’s a solution to the second equation. We’ll say that (2,3) is a solution to this system of linear equations, since it is a solution to each of the equations. Next Slide

  3. Practice Problems • Is (4,0) a solution to the first equation (in the system of equations discussed on the previous slide (slide 2))? • Is (4,0) a solution to the second equation? • Is (4,0) a solution to the system of equations? Click for answers: 1) Yes; 2) No; 3) No, to be a solution to a system of equations, you have to be a solution to all the equations. Next Slide

  4. Method I (Substitution) We’ll look at two basic methods for finding solutions to a system of equations. The first is called substitution. For the system 3x + 2y = 12 and y = 3x –3, we start by assuming that we have a solution (x,y) that satisfies both equations, and then we figure out what those numbers must be. If 3x + 2y and 12 are equal, and y and 3x – 3 are also equal, then y and 3x – 3 are interchangeable, and we can replace y by 3x –3 in the first equation The two sides must still be equal. We must have 3x + 2(3x – 3) = 12. (cont.) Next Slide

  5. Example (cont.) When we substituted the second equation into the first, we got 3x + 2(3x – 3) = 12. We’re down to one variable now (which is good, and always the goal), and we now have linear equations in one variable, which we already know how to solve. 3x + 6x – 6 = 12, 9x = 18, x = 2. We now know the x-coordinate of the solution. To find y, we can substitute back into any of the original equations. Substituting into the second equation, for example, y = 3(2) – 3, and y = 6 – 3 = 3. The solution, therefore, is (x,y) = (2,3) (which we already knew). Next Slide

  6. Method I Overview • Solve one equation for one of the variables. (It doesn’t matter which equation or which variable.) • Substitute for that variable in the other equation. (This should get you down to one variable). • Solve for that variable. • Substitute this value into one of the original equations to find the other variable. Next Slide

  7. Example Consider the system of linear equations 2x – 5y = 18 and x = 3y + 10. The second equation is already solved for x. Substitute into first. 2(3y + 10) – 5y = 18 Solve for y. 6y + 20 – 5y = 18 y = 2. Substitute into one of the original equations to find x. In the second equation: x = 3(-2) + 10, x = 6 + 10 = 4. Solution: (4,-2). Next Slide

  8. Example Solve the system 5x – y = 8 and 3x + 2y = 3. It looks easiest to solve the first equation for y. 5x – y = 8, y = 5x + 8, y = 5x – 8. Substitute into the second equation, and solve for x. 3x + 2(5x – 8) = 3, 3x + 10x – 16 = 3, 13x = 13, x = 1. Substitute into y = 5x – 8 (it is equivalent to the first equation), y = 5(1) – 8, y = 5 – 8 = 3. Solution: (1,-3). Next Slide

  9. Practice Problems • Solve the system y = 2x + 1 and 3x – 4y = 9. • Solve the system 2x – 3y = 7 and 2x + y = 3. (I would solve the second equation for y). Click for answers: 1) (1,3); 2) (2,-1). Next Slide

  10. Method II (Elimination, a.k.a., Addition) Example, 3x + 2y = 19 and x – 2y = 1. This system tells us that 3x + 2y and 19 are equal, As are x – 2y and 1. Therefore, if we add 3x + 2y and x – 2y together (the left sides), Then this should be equal to 19 plus 1 (the right sides). Combining like terms gives us 4x + 0 = 20. We essentially just added the two equations together. And we got down to one variable, which is very good. Let’s start over, and write it as we generally will. Next Slide

  11. Example (cont.) 3x + 2y = 19 x – 2y = 1 Add to get: 4x + 0 = 20. Solve for x: 4x = 20, and then x = 5. Now finish up as in Method I. Substitute into one of the equations. (5) – 2y = 1. 2y = 4 y = 2. Solution: (5,2). Next Slide

  12. Method II Overview We can always add two equations together. In this last example, we got something good, because one variable went away. Since we can also always multiply an equation by a constant, we can set things up so that one variable will cancel. • Put both equations into the form: x-term plus y-term equals constant term, and write them so that like terms are in their own columns. • Multiply one or both equations by a number so that one variable will cancel out after adding. • Add. • Solve for the one remaining variable. • Substitute into one of the equations to get the other variable. Next Slide

  13. Example 3x – 2y = 19 x + 5y = 12. If we multiply the second equation by 3, we’ll get 3x in the first and –3x in the second, and they will cancel. 3x – 2y = 19 3x – 15y = 36 Add: 0 – 17y = 17, 17y = 17, y = 1. Substitute: x + 5(1) = 12, x + 5 = 12, x = 7. Solution: (7,1). Next Slide

  14. Example 2x + 5y = 11 3x – 4y = 18 We can do this several ways. One way is to get 6x in the first equation and –6x in the second. We can do this by multiplying by 3 and 2. 6x + 15y = 33 6x + 8y = 36 Add: 23y = 69, y = 3. Substitute: 2x + 5(3) = 11, 2x + 15 = 11, 2x = 4, and x = 2. Solution: (-2,3). Next Slide

  15. Practice Problems • Solve by elimination (or adding) 2x + 3y = 7 and 2x + y = 3. • Solve 2x + 3y = 14 and 5x – y = 1. • Solve 3x – 4y = 17 and 2x – 3y = 12. Click for answers: 1) (-2,-1); 2) (1,4); 3) (-3,2). Next Slide

  16. Special Cases. Consider this example. 4x + 2y = 6 and y = -2x + 1. By substitution, we get: 4x + 2(-2x + 1) = 6, 4x – 4x + 2 = 6, and 2 = 6. Something is obviously wrong. What we can see here is that 2 = 6 cannot ever be true. Therefore, there can be no solution. Look at the graphs of the two lines. Next Slide

  17. Special Cases (cont.) The solution to a system of equations is a point that lies on both lines. In this last example, the lines are parallel, so no point can lie on both lines. Another example. Consider 4x + 2y = 2 and y = 2x + 1. Again by substitution: 4x + 2(-2x +1) = 2, 4x – 4x + 2 = 2, and 0 = 0. We see a problem again, but here the equation is always true. Look at the graph of the two equations. Next Slide

  18. Special Cases (cont.) In this last example, the two lines are actually the same line, so every point on the line is a solution to the system. You can recognize the two special cases as follows. If in solving the system, you get something like 2 = 4, which is not true, then the lines are parallel (and different), and there are no solutions. If in solving the system, you get something like 0 = 0, which is always true, then the lines are the same, and there are infinitely many solutions. Next Slide

  19. Practice Problems • Solve 2x – 3y = 9 and y = (2/3)x + 2. • Solve 4x + 2y = 2 and 2x – y = 1. Click for solutions: 1) No solutions. 2) Infinitely many solutions. End

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