# MATH 1010 Sec. 3 Quiz 5 February 17, 2011

```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&middot;
Equation 2:
2&middot;
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
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