Algebra 5
5 × equation (1) gives:
35x − 10y = 130
(3)
2 × equation (2) gives:
12x + 10y = 58
(4)
equation (3) +equation (4) gives:
47x + 0 = 188
from which,
x =
188
47
= 4
Substituting x = 4 in equation (1) gives:
28 − 2y = 26
from which, 28 − 26 = 2y and y = 1
Problem 19. Solve
x
8
+
5
2
= y
(1)
11 +
y
3
= 3x.
(2)
8 × equation (1) gives: x + 20 = 8y
(3)
3 × equation (2) gives: 33 + y = 9x
(4)
i.e.
x − 8y = −20
(5)
and
9x − y = 33
(6)
8 × equation (6) gives: 72x − 8y = 264
(7)
Equation (7) − equation (5) gives:
71x = 284
from which,
x =
284
71
= 4
Substituting x = 4 in equation (5) gives:
4 − 8y = −20
from which,
4 + 20 = 8y and y = 3
(d) Quadratic equations
Problem 20. Solve the following equations by
factorization:
(a) 3x 2 − 11x − 4 = 0
(b) 4x 2 + 8x + 3 = 0.
(a) The factors of 3x 2 are 3x and x and these are placed
in brackets thus:
(3x
)(x
)
The factors of −4 are +1 and −4 or −1 and
+4, or −2 and +2. Remembering that the product of the two inner terms added to the product
of the two outer terms must equal −11x, the only
combination to give this is +1 and −4, i.e.,
3x 2 − 11x − 4 = (3x + 1)(x − 4)
Thus (3x + 1)(x − 4) = 0 hence
either
(3x + 1) = 0 i.e. x = −
1
3
or
(x − 4) = 0 i.e. x = 4
(b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1)
Thus (2x + 3)(2x + 1) = 0 hence
either
(2x + 3) = 0 i.e. x =−
3
2
or
(2x + 1) = 0 i.e. x = −
1
2
Problem 21. The roots of a quadratic equation
are
1
3 and −2. Determine the equation in x.
If
1
3 and −2 are the roots of a quadratic equation then,
(x −
1
3 )(x + 2) = 0
i.e. x
2
+ 2x −
1
3 x −
2
3 = 0
i.e.
x 2 +
5
3 x −
2
3 = 0
or
3x 2 + 5x −2 = 0
Problem 22. Solve 4x 2 + 7x + 2 = 0 giving the
answer correct to 2 decimal places.
From the quadratic formula if ax 2 + bx + c = 0 then,
x =
−b ±
√
b 2 − 4ac
2a
Hence if 4x 2 + 7x + 2 = 0
then x =
−7 ±
7 2 − 4(4)(2)
2(4)
=
−7 ±
√
17
8
=
−7 ± 4.123
8
=
−7 + 4.123
8
or
−7 − 4.123
8
i.e. x = −0.36 or −1.39
5 × equation (1) gives:
35x − 10y = 130
(3)
2 × equation (2) gives:
12x + 10y = 58
(4)
equation (3) +equation (4) gives:
47x + 0 = 188
from which,
x =
188
47
= 4
Substituting x = 4 in equation (1) gives:
28 − 2y = 26
from which, 28 − 26 = 2y and y = 1
Problem 19. Solve
x
8
+
5
2
= y
(1)
11 +
y
3
= 3x.
(2)
8 × equation (1) gives: x + 20 = 8y
(3)
3 × equation (2) gives: 33 + y = 9x
(4)
i.e.
x − 8y = −20
(5)
and
9x − y = 33
(6)
8 × equation (6) gives: 72x − 8y = 264
(7)
Equation (7) − equation (5) gives:
71x = 284
from which,
x =
284
71
= 4
Substituting x = 4 in equation (5) gives:
4 − 8y = −20
from which,
4 + 20 = 8y and y = 3
(d) Quadratic equations
Problem 20. Solve the following equations by
factorization:
(a) 3x 2 − 11x − 4 = 0
(b) 4x 2 + 8x + 3 = 0.
(a) The factors of 3x 2 are 3x and x and these are placed
in brackets thus:
(3x
)(x
)
The factors of −4 are +1 and −4 or −1 and
+4, or −2 and +2. Remembering that the product of the two inner terms added to the product
of the two outer terms must equal −11x, the only
combination to give this is +1 and −4, i.e.,
3x 2 − 11x − 4 = (3x + 1)(x − 4)
Thus (3x + 1)(x − 4) = 0 hence
either
(3x + 1) = 0 i.e. x = −
1
3
or
(x − 4) = 0 i.e. x = 4
(b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1)
Thus (2x + 3)(2x + 1) = 0 hence
either
(2x + 3) = 0 i.e. x =−
3
2
or
(2x + 1) = 0 i.e. x = −
1
2
Problem 21. The roots of a quadratic equation
are
1
3 and −2. Determine the equation in x.
If
1
3 and −2 are the roots of a quadratic equation then,
(x −
1
3 )(x + 2) = 0
i.e. x
2
+ 2x −
1
3 x −
2
3 = 0
i.e.
x 2 +
5
3 x −
2
3 = 0
or
3x 2 + 5x −2 = 0
Problem 22. Solve 4x 2 + 7x + 2 = 0 giving the
answer correct to 2 decimal places.
From the quadratic formula if ax 2 + bx + c = 0 then,
x =
−b ±
√
b 2 − 4ac
2a
Hence if 4x 2 + 7x + 2 = 0
then x =
−7 ±
7 2 − 4(4)(2)
2(4)
=
−7 ±
√
17
8
=
−7 ± 4.123
8
=
−7 + 4.123
8
or
−7 − 4.123
8
i.e. x = −0.36 or −1.39