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MATH 10 Solving Systems of Linear Equations What is a SYSTEM OF EQUATIONS ? It’s any 2 or more equations considered together. To solve a system of linear equations, we will find the point where the graphs of the two linear equations intersect. A. Solve by GRAPHING 1) Where do the graphs of y = 2x + 1 and y = −3x + 6 intersect? Step 1: draw the graph of y = 2x + 1 by marking Step 2: draw the graph of y = −3x + 6 by marking 2 3 the y-intercept at (0,1) and making slope . the y-intercept at (0,6) and making slope − . 1 1 5 .0 5.0 -5.0 5.0 -5.0 5.0 You can see clearly that the lines cross at (1,3). This is the solution of the system. 2) Solve the system x + 3y = 8 2y + 3x + 4 = 0 Try putting each equation in slope-intercept form to make them easier to graph. x + 3y = 8 2y + 3x + 4 = 0 −x −x − 3x − 4 −3x − 4 3y − x + 8 2y −3x − 4 = 3 = 3 2 2 1 8 3 y = − 3 x + 3 (still ugly!) y = − 2 x − 2 (easy to graph!) If slope-intercept form is still ugly, try getting one convenient ordered pair. 1 8 3 e.g. if x = 5, y = − 3(5) + 3 = 3 = 1 5.0 1 Then make slope − 3 from that point. You can see that the solution of the system is (−4,4). -5.0 5.0 B. Solve by ADDITION If the two lines don’t meet at a convenient ordered pair, solving by GRAPHING can be very inconvenient. ADDITION (or Linear-Combination) is the name given to one algebraic method of solving systems. 1) The sum of 2 numbers is 6. If the smaller number is subtracted from the larger one, the difference is 10. Find the numbers. Call the numbers x and y. The first sentence is translated as x + y = 6. Pick one of them to be the smaller number. If y is smaller, then the second sentence is translated as x − y = 10 To understand the “addition”, watch this: If = So if x + y = 6 and = and x − y = 10 then + = + then 2x + 0 = 6 + 10 2x = 16 x=8 Then, since x + y = 6, we have 8 + y = 6, so y = −2. The graphs of the two equations would cross at (8,−2). The answer to the problem is “the 2 numbers are 8 and −2”. (Note: check and you’ll see that these work in the 2nd equation too.) 2) 3x + 4y = 6 3) 2x − y − 6 = 0 −3x + 3y = −20 5y − 2x = 10 ADD: 0 + 7y = −14 ADD: 0 + 4y − 6 = 0 + 10 y = −2 +6 +6 4y = 16, so y = 4 Then 3x + 4(−2) = 6 3x − 8 = 6 Then 2x − 4 − 6 = 0 +8 +8 2x − 10 = 0 3x 14 3 = 3 + 10 + 10 2x = 10, so x = 5 14 − Solution: ( 3 ,−2) Solution: (5,4) NOTE: When you ADDED, one variable was eliminated. What if this doesn’t happen? 4) 2x − 5y = −17 If you ADD, you get 8x + 2y = −2 and you still have 2 variables. 6x + 7y = 15 Instead, multiply both sides of the first equation by −3. Then you have −3(2x − 5y) = (−17)−3 --------> −6x + 15y = 51 and the second equation ---------------> 6x + 7y = 15 Now ADD 0 + 22y = 66 x’s are eliminated y = 3 Then 2x − 5(3) = −17 2x − 15 = −17 + 15 + 15 2x = −2, so x = −1 − Solution: (−1,3) 5) Sometimes you need to multiply in both equations so that one variable is eliminated. 5x − 4y = 14 multiply both sides by 2: 2(5x − 4y) = (14)2 ----> 10x − 8y = 28 2x + 3y = −22 multiply both sides by −5: −5(2x + 3y) = (−22)(−5) ---> −10x − 15y = 110 Now ADD -------------> 0 − 23y = 138 y = −6 Then 5x − 4(−6) = 14 5x + 24 = 14 − 24 −24 5x = −10, so x = −2 Solution: (−2, −6) SUMMARY OF STEPS FOR “ADDITION” - multiply one or both equations so that one variable is eliminated i.e., so that the x’s add up to zero, or the y’s add up to zero - ADD the equations on the left side, on the right side - solve the equation for the variable - plug the value back into one of the original equations (Note: it doesn’t matter which one.) - write the solution as an ordered pair (x,y) or give the answer in “words”. C. Solve by SUBSTITUTION 1) 7x + y = 13 1st equation has just one y, which you can easily isolate: y = 13 − 7x 2x − 3y = 30 Since we are looking for the y-value that is the same in both equations, we can SUBSTITUTE our isolated y into the second equation, as shown: 2x − 3(13 − 7x) = 30 Then solve for x: 2x − 39 + 21x = 30 23x − 39 = 30 + 39 + 39 23x = 69 x = 3 Then 7(3) + y = 13 21 + y = 13 −21 −21 y = −8 − Solution: (3,−8) 2) 3x − 8y = −4 x + 6y = 16 Now it’s easiest to isolate x , in the 2nd equation: x = 16 − 6y SUBSTITUTE into the first equation: 3(16 − 6y) − 8y = −4 48 − 18y − 8y = −4 −48 −48 − 26y = −52 , so y = 2 Then 3x − 8(2) = −4 3x − 16 = −4 + 16 + 16 3x = 12 , so x = 4 Solution: (4,2) SUMMARY OF STEPS FOR “SUBSTITUTION” - find one equation where it’s easy to isolate one variable (usually where you have just one x or one y) - substitute into the other equation - solve the equation for the variable - plug the value back into one of the original equations (Note: it doesn’t matter which one.) - write the solution as an ordered pair (x,y) or give the answer in “words”. D. Solve by COMPARISON 1) 6x + 1 + 11y = 0 Look for something that is the SAME in both equations. 11y + 5 = −2x Do you see they both have 11y ? Isolate 11y in both equations. 4 11y + 5 1st equation: 11y = −6x − 1 2nd equation: 4( 4 ) = (−2x)4 11y + 5 = −8x 11y = −8x − 5 COMPARE the 11y in the 1st equation, the 11y in the 2nd equation – they must be equal! −6x − 1 = −8x − 5 + 8x + 8x In practice is would have been 2x − 1 = −5 easier to simply substitute for +1 +1 “11y” in the second equation… 2x = −4, so x = −2 Think of the COMPARISON method as the SUBSTITUTION Then 6(−2) + 1 + 11y = 0 method used twice! −12 + 1 + 11y = 0 −11 + 11y = 0 + 11 + 11 11y = 11 , so y = 1 − Solution: (−2,1) 2) The sum of 4 times a number and 7 times a second number is 14. If double the first number is subtracted from 7 times the second number, the result is −10. Find the two numbers. First sentence: 4x + 7y = 14 Second sentence: 7y − 2x = −10 Notice 7y in both equations. Isolate 7y in 1st equation: 7y = 14 − 4x Isolate 7y in 2nd equation: 7y = 2x − 10 Therefore it must be true that 14 − 4x = 2x − 10 + 4x + 4x 14 = 6x − 10 + 10 + 10 24 = 6x so x = 4 Then 4(4) + 7y = 14 16 + 7y = 14 − 16 − 16 2 2 7y = −2 so y = − 7 Answer: The numbers are 4 and −7. SUMMARY OF STEPS FOR “COMPARISON” - find something the same in both equations - isolate for what is the same - write the two equations equal to each other - solve for the variable - plug the value back into one of the original equations (Note: it doesn’t matter which one.) - write the solution as an ordered pair (x,y) or give the answer in “words”.