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MATH 1010 Sec. 3 Quiz 5 February 17, 2011 1. Solve the following system of equations 2x − y = 4 2x − 3y = 2 Subtracting Equation 2 from Equation 1 we get 2y = 2 Dividing by 2 it becomes y=1 Substituting the value of y in Equation 1 we get 2x − 1 = 4 Therefore x = 25 . Solution = ( 52 , 1). Check: (optional) Equation 1: 5 −1=4 2 5−1=4 2· Equation 2: 2· 5 − 3(1) = 2 2 5−3=2 2. Find two positive integers such that the sum of the numbers is 18 and their difference is 4. We have to solve the following system n + m = 18 n−m=4 Summing Equation 2 and Equation 1 we get 2n = 22 Dividing by 2 it becomes n = 11 Substituting the value of n in Equation 1 we get 11 + m = 18 Therefore m = 7. In conclusion the two integers are 7 and 11. 1 3. Solve the following system of equations x + 2y − z = 2 2x + 3y = 4 y+z =3 Computing: 2 (Equation 1) − Equation 2 and substituting in Equation 2 we get x + 2y − z = 2 y − 2z = 0 y+z =3 Now computing Equation 3 − Equation 2, we get 3z = 3. Therefore z = 1. Substituting z = 1 in y − 2z = 0, we get y = 2. Finally, substituting z = 1 and y = 2 in x + 2y − z = 2, we get x = −1. The solution is (−1, 2, 1). 2