Download MATH 1010-2: Quiz 5A

MATH 1010-2: Quiz 5A
Solution
1. Let f (x) = x2 . Evaluate
(a) (2 points) f (−1)
Solution: f (−1) = (−1)2 = 1
(b) (3 points) f (a + 1)
Solution: f (a + 1) = (a + 1)2 = a2 + 2a + 1
2. (5 points) Solve the following system of equations by the method of substitution or elimination.
(
2x − 3y = 1
4x − 5y = 3
Solution: From the first equation, we have
x=
3
1
y+
2
2
Substitute x in the second equation, we have
3
1
4 · ( y + ) − 5y = 3
2
2
Solve it and get
y=1
Back-substitute y in
x=
we have
x=
To conclude, (x, y) = (2, 1)
3
1
y+
2
2
3
1
·1+ =2
2
2
MATH 1010-2: Quiz 5B
Solution
1. Let f (x) = x2 . Evaluate
(a) (2 points) f (−2)
Solution: f (−2) = (−2)2 = 4
(b) (3 points) f (b − 1)
Solution: f (b − 1) = (b − 1)2 = b2 − 2b + 1
2. (5 points) Solve the following system of equations by the method of substitution or elimination.
(
3x − 2y = −1
5x − 4y = −3
Solution: From the first equation, we have
y=
3
1
x+
2
2
Substitute y in the second equation, we have
3
1
5x − 4 · ( x + ) = −3
2
2
Solve it and get
x=1
Back-substitute x in
y=
we have
y=
To conclude, (x, y) = (1, 2)
3
1
x+
2
2
3
1
·1+ =2
2
2