The given system of equations are 8x − 3y = 5xy ...(1)
6x − 5y = −2xy ...(2)
Observe that the given system is not linear because of the occurrence of xy term. Also note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other solution would have both x \(\neq \) 0 and y \(\neq \) 0
Let us take up the case where x \(\neq \) 0 and y \(\neq \) 0
Dividing both sides of each equation by xy,
\(\frac { 8x }{ xy } -\frac { 3y }{ xy } =\frac { 5xy }{ xy } \)
\(\frac { 6x }{ xy } -\frac { 5y }{ xy } =\frac { -2xy }{ xy } \)
Let \(a=\frac { 1 }{ x } ,b=\frac { 1 }{ y } \)
We get, \(\frac { 8 }{ y } -\frac { 3 }{ x } =5\)  .....(3)
\(\frac { 6 }{ y } -\frac { 5 }{ x } =-2\)  .......(4)
(3)&(4) respectively become, 8b − 3a = 5 ...(5)
b − 5a = −2 ...(6)
which are linear equations in a and b.
To eliminate a, we have,  (5) \(\times\) 5\(\Rightarrow\) 40b −15a = 25 .....(7)
(6) × 3\(\Rightarrow\) 18b −15a = −6 .....(8)
Now proceed as in the previous example to get the solution\(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \).
Thus, the system have two solutions \(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \) and (0,0).

We have already come across one method: forming a table of values, listing and plotting ordered pairs and joining the points.
But, to fix a line, after all, how many points do we need? Just two! These can easily be obtained when a line is given in the form y = mx + c.
The given line y = 4x – 3
put x = 0 to get y-intercept
y = 4(0)–3
y = –3
point is (0, –3) and y-intercept = – 3
put y = 0 to get x-intercept
0 = 4x – 3
3 = 4x
\(\frac { 3 }{ 4 } =x\)
point is \(\left( \frac { 3 }{ 4 } ,0 \right) \) and x-intercept = \(\frac { 3 }{ 4 } \)
The graph may be drawn through two points (0,−3) and \(\left( \frac { 3 }{ 4 } ,0 \right) \)

(i) x³ - 3x² -10x + 24
Let p(x) = x³ - 3x² -10x + 24
Sum of all the co-efficients = 1- 3 - 10 + 24 = 25 -13 = 12 ≠ 0
Hence (x - 1) is not a factor.
Sum of co-efficient of even powers with constant = -3 + 24 = 21
Sum of co-efficients of odd powers = 1 - 10 = - 9
21 ≠ -9
Hence (x + 1) is not a factor.
p(2) = 2³+ 3(2²) -10\(\times\)2+24
= 8 -12 - 20 + 24
= 32 - 32 = 0 ஃ (x - 2) is a factor.
Now we use synthetic division to find other factor

Thus (x - 2) (x + 3) (x - 4) are the factors.
ஃ x³ - 3x² - 10x + 24 = (x - 2)(x + 3)(x - 4)
(ii) 2x² - 3x² - 3x + 2
Let p (x) = 2x³ - 3x² - 3x + 2
Sum of all the co-efficients are
2 - 3 - 3 + 2 = 4 - 6 = -2 ≠ 0
ஃ (x - 1) is not a factor.
Sum of co-efficients of even powers of x with constant = -3 + 2 = - 1
Sum of co-efficients of odd powers of x = 2 - 3 = - 1
(-1) = (-1)
ஃ (x + 1) is a factor
Let us find the other factors using synthetic division

Quotient is 2x²- 5x + 2 = 2x - 4x - x + 2 = 2x (x - 2) - 1 (x - 2)
= (x - 2)(2x - 1)
ஃ 2x³- 3x²- 3x +2 = (x +1) (x - 2) (2x - 1)
(iii) -7x+ 3 + 4x³
Letp (x) = 4x³ + 0x² - 7x + 3
Sum of the co-efficients are = 4 + 0 - 7 + 3
= 7 - 7 = 0
ஃ (x - 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 3 = 3
Sum of co-efficients of odd powers of x with constant = 4 - 7 = -3
3 ≠ -3
ஃ (x + 1) is not a factor
Using synthetic division, let us find the other factors.


Quotient is 4x² + 4x - 3
= 4x²+ 6x - 2x - 3
= 2x(2x + 3) -1(2x + 3)
= (2x + 3)(2x - 1)
ஃ The factors are (x -1), (2x + 3) and (2x - 1)
ஃ -7x + 3 + 4x³= (x+1)(2x+3)(2x-1)
(iv) x³ + x² - 14x - 24
Let p(x) = x³ + x² - 14x - 24
Sum of the co-efficients are 1 + 1-14 - 24 = -36 ≠ 0
ஃ (x - 1) is not a factor
Sum of co-efficients of even powers of x with constant = 1 - 24 = -23
Sum of co-efficients of odd powers of x = 1 - 14 = -3
-23 ≠ -13
ஃ (x + 1) is also not a factor
p(2) = 2³ + 2² - 14 (2) - 24 = 8 + 4 - 28 - 24
= 12 - 52 ≠ 0, (x - 2) is a not a factor
p(-2) = (-2)³ + (-2)² - 14 (-2) - 24
-8 + 4 + 28 - 24 = 32 - 32 = 0
ஃ (x + 2) is a factor
To find the other factors let us use synthetic division
x³+x²-14x-24


ஃ The factors are (x + 2),(x + 3),(x - 4)
ஃ x³+x²-14x - 24 = (x + 2)(x + 3)(x - 4)
(v) x³- 7x + 6
Let p (x) = x³+ 0x²-7x + 6
Sum of the co-efficients are = 1+0 -7+ 6 = 7-7 = 0
ஃ (x - 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 6 = 6
Sum of coefficient of odd powers of x = 1 - 7 = - 7
6 ≠ -7
ஃ (x + 1) is not a factor
To find the other factors, let us use synthetic division.


ஃ The factors are (x - 1), (x - 2), (x + 3)
ஃ x³ + 0x² - 7x + 6 = (x - 1)(x - 2)(x + 3)
(vi) x³-10x²- x + 10
Let p(x) = x³-10x²- x + 10
Sum of the co-efficients = 1 - 0 - 1 + 10
= 11 -11 = 0
(x - 1) is a factor
Sum of co-efficients of even powers of x with constant = -10 + 10 = 0
Sum of co-efficients of odd powers of = 1 - 1 = 0
ஃ (x + 1) is a factor
Synthetic division

ஃ x³ + 10x²- x + 10 = (x - 1) (x + 1)(x -10)

(i) x²+10x + 24 = x²+ 6x + 4x + 24
= x(x + 6) + 4(x + 6)
= (x + 6)(x + 4)

(ii) z²+ 4z -12
= z²+ 6z - 2z -12
= z(z + 6) -2(z + 6)
= (z + 6)(z - 2)

(iii) p²- 6p -16 = p²- 8p + 2p -16
= p(p - 8)+2(p - 8)
= (p - 8)(p + 2)

(iv) t²+ 72 -17t = t²-17t + 72
= t²- 9t - 8t + 72
= t(t - 9) - 8 (t - 9)
= (t - 9)(t - 8)

(v) y² -16y- 80 = y² - 20y + 4y - 80
= y(y - 20) + 4 (y - 20)
= (y - 20) (y + 4)

(vi) a² + 10a - 600
= a² + 30a - 20a - 600
= a(a + 30)- 20(a + 30)
= (a + 30)(a - 20)
​​​​​

Compare with ax²+bx+c
we get, a = 2, b = -15, c = 27
product ac = 2\(\times\) 27 = 54 and sum b = -15
∴ we split the middle term as -6x and -9x

The required factors are -6 and -9
2x²-15x + 27 = 2x²- 6x - 9x + 27
=2x(x - 3) -9(x - 3)
=(x - 3)(2x - 9)
Therefore, (x - 3) and (2x - 9) are the factors of 2x²-15x + 27

(i) 16x³y² = \(2 \times 2 \times 2 \times 2 \times x^{3} y^{2}=2^{4} \times x^{3} \times y^{2}=2^{3} \times 2 \times x^{2} \times x \times y^{2}\)
24xy³z = \(2 \times 2 \times 2 \times 3 \times x \times y^{3} \times z=2^{3} \times 3 \times x \times y^{3} \times z=2^{3} \times 3 \times x \times y \times y^{2} \times z\)
Therefore, GCD = 2³xy²
(ii) y³+1 = y³+1³ = (y+1)(y²-y+1)
y²-1=y²-1² = (y+1)(y-1)
Therefore, GCD = (y+1)
(iii) 2x²-18 = 2(x²-9) = 2(x²-3²) = 2(x+3)(x-3)
x²-2x-3 = x²-3x+x-3
= x(x-3)+1(x-3)
= (x-3)(x+1)
Therefore, GCD = (x-3)
(iv) (a-b), (b-c)³, (c-a)⁴
There is no common factor other than one.
Therefore, GCD = 1

(i) am+bm+cm
am+bm+cm
m(a+b+c) factored form
(ii) a³-a²b
a². a - a². b group in pairs
a² \(\times\) (a-b) factored form
(iii) 5a −10b −4bc +2ac
5a-10b+2ac-4bc
5(a-2b)+2c(a-2b)
(a-2b)(5+2c)
(iv) x + y −1−xy
x-1+y-xy
(x-1)+y(1-x)
(x-1)-y(x-1)   [(a-b)= -(b-a)
(x-1)(1-y)

(i) 98² = (100-2)³
(a-b)³ ≡ a³- 3a²b + 3ab²- b³
98³ = (100 - 2)² = 100³- 3\(\times\)100²\(\times\)2 + 3\(\times\)100\(\times\)2²- 2³
= 1000000 - 3\(\times\)10000\(\times\)2 + 300\(\times\)4 - 8
= 1000000 - 60000 +1200 - 8 =1001200 - 60008
= 941192
(ii) 1001³= (1000 + 1)³
(a+b)³ ≡ a³+3a²b+3ab²+b³
(1000+1)³=1000³+3(1000)\(\times\)1²+1³
= 1000,000,000 + 3,000,000 + 3000 + 1
= 1,003,003,001

(x + a)(x + b)(x + c) = x³+14x²+ 59x + 70...(1)
(x + a)(x + b)(x + c) ≡ x³+ (a + b + c)x²+ (ab + bc + ca)x + abc...(2)
(i) Comparing (1) and (2)
We get, a + b + c =14
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=\frac{59}{70}\)
(iii) (a + b + c)² = a²+ b²+ c² + 2 ab + 2 bc + 2 ca
a²+ b²+ c² = (a+b+c)²- 2(ab+bc+ca)
= 14²-2(59) =196-118 = 78
(iv) \(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}=\frac{78}{70}=\frac{39}{35}\)