1. Solve these equations…
a)
6a + 9 = a – 5
2. Simplify fully…
a)
5a3b4 x 5ab3 =
b)
8a7 =
24a3
(4a3)3 =
b)
a4 = 625
c)
(3a – 1)(a + 3) = 0
c)
d)
a2 + 9a + 14 = 0
3. Factorise…
a)
e)
a2 – 36 =
2a(9 – a) = 0
f) Jim buys some fish (f) and
some scoops of chips (c). Jim
spends a total of $23.00. Jim
writes an equation for the amount
he spends…
2a2 + 16a + 32 =
5. Simplify fully…
ii) Jim bought one more fish than
he did scoops of chips. How many
fish did he buy in total?
7. Write an expression for the
perimeter of the following shape…
3a
31cm
4. Rearrange A = πr2 to make
r the subject of the formula.
2.6f + 2.5c = 23
i) explain each part of this
equation.
a + 3b = 45
2a + b = 30
a
b)
6. Simultaneously solve to find
the solutions for both a and b
46cm
Yr 11 MCAT Algebra Practice
a)
b)
7a3 x 5ab5 =
25b3 14a2
a2 + 13a + 36 =
a+9
8. If the total perimeter of the
above was 226cm what is the value
of a?
Yr 11 MCAT Algebra Practice
1. Solve these equations…
c)
5a = - 14
a = -14/5 -24/5 or -2.8
a4 = 625
b)
a = 4√625
a = 5 and -5
5a3b4 x 5ab3 = 25a4 b7
8a7 = a4
3
24a3
e)
eqn 1
eqn 2
double eqn 1
subtract the two equations
5b = 60
b = 12
substitute b into equation 2
(3a – 1)(a + 3) = 0
a2 + 9a + 14 = 0
( a + 7 )( a + 2 ) = 0
a is ____
-2
-7 or ____
c)
2a + 12 = 30
2a = 18
a=9
(4a3)3 = 64 a9
factorise first
7. Write an expression for the
perimeter of the following shape…
3a
must have both
3. Factorise…
a)
a2 – 36 = ( a – 6 )(a + 6 )
the difference of two squares
2a(9 – a) = 0
9
a is ____
0 or ____
b)
2a2 + 16a + 32 = 2(a2 + 8a + 16)
take out common factor
4. Rearrange A = πr2 to make
r the subject of the formula.
πr2 = A
r2 = A/π
r = ±√(A/π)
2.6f + 2.5c = 23
2.5 is the price of the chips.
c is the number of scoops.
i) explain each part of this
equation. 23 is the total amount he paid
ii) Jim bought one more fish than
he did scoops of chips. How many
fish did he buy in total? five fish
= 2(a + 4)(a + 4)
can be 2(a + 4)2
5. Simplify fully…
a)
1
5
b)
1
3a + 31
perimeter = all sides added
perimeter = 46+3a+46–a+31+a+3a+31
perimeter = 6a + 154
8. If the total perimeter of the
above was 226cm what is the value
of a?
x 5ab5 = a4b5 = a2b2
10a2b3
10
25b3 14a2
7a3
2
a2 + 13a + 36 =
a+9
31cm
a
f) Jim buys some fish (f) and
some scoops of chips (c). Jim
spends a total of $23.00. Jim
writes an equation for the amount
he spends…
2.6 is the price of the fish.
f is the number of the fish.
a + 3b = 45
2a + b = 30
.
1/
-3
a is ____
or ____
3
d)
2a + 6b = 90
46 – a
b)
6a + 9 = a – 5
a)
6. Simultaneously solve to find
the solutions for both a and b
46cm
a)
2. Simplify fully…
(a + 9 )( a + 4) = a + 4
a+9
6a + 154 = 226
6a = 72
a = 12cm