Question

cross multiply the equation :
x+3/x-3 + x+2/x-2=2

Answer 1

Here we will discuss about simultaneous linear equations by using cross-multiplication method.

General form of a linear equation in two unknown quantities: 

ax + by + c = 0, (a, b ≠ 0) 

Two such equations can be written as: 

a₁x + b₁y + c₁ = 0 ----------- (i) 

a₂x + b₂y + c₂ = 0 ----------- (ii) 

Let us solve the two equations by the method of elimination, multiplying both sides of equation (i) by a₂ and both sides of equation (ii) by a₁, we get: 

a₁a₂x + b₁a₂y + c₁a₂ = 0

a₁ a₂x + a₁b₂y + a₁c₂ = 0

Subtracting, b₁a₂y - a₁b₂y + c₁a₂ - c₂a₁ = 0

or, y(b₁ a₂ - b₂a₁) = c₂a₁ - c₁a₂



 

Therefore, y = (c₂a₁ - c₁a₂)/(b₁a₂ - b₂a₁) = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0

Therefore, y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁), ------------- (iii) 

Again, multiplying both sides of (i) and (ii) by b₂ and b₁ respectively, we get; 

a₁b₂x + b₁b₂y + b₂c₁ = 0

a₂b₁x + b₁b₂y + b₁c₂ = 0

Subtracting, a₁b₂x - a₂b₁x + b₂c₁ - b₁c₂ = 0

or, x(a₁b₂ - a₂b₁) = (b₁c₂ - b₂c₁)

or, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

Therefore, x/(b₁c₂ - b₂c₁) = 1/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0 -------------- (iv) 

From equations (iii) and (iv), we get: 

x/(b₁c₂ - b₂c₁) = y/(c₁a₂) - c₂a₁ = 1/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0

This relation informs us how the solution of the simultaneous equations, co-efficient x, y and the constant terms in the equations are inter-related, we can take this relation as a formula and use it to solve any two simultaneous equations. Avoiding the general steps of elimination, we can solve the two simultaneous equations directly. 

So, the formula for cross-multiplication and its use in solving two simultaneous equations can be presented as: 

If (a₁b₂ - a₂b₁) ≠ 0 from the two simultaneous linear equations

a₁x + b₁y + c₁ = 0 ----------- (i) 

a₂x + b₂y + c₂ = 0 ----------- (ii) 

we get, by the cross-multiplication method: 

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) ---------- (A) 

That means, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)

Note:

If the value of x or y is zero, that is, (b₁c₂ - b₂c₁) = 0 or (c₁a₂ - c₂a₁) = 0, it is not proper to express in the formula for cross- multiplication, because the denominator of a fraction can never be 0. 

From the two simultaneous equations, it appears that the formation of relation (A) by cross-multiplication is the most important concept. 

At first, express the co-efficient of the two equations as in the following form: 





Now multiply the co-efficient according to the arrow heads and subtract the upward product from the downward product. Place the three differences under x, y and 1 respectively forming three fractions; connect them by two signs of equality. 
Worked-out examples on simultaneous linear equations by using cross-multiplication method: 


1. Solve the two variables linear equation: 

8x + 5y = 11

3x – 4y = 10

Solution:

On transposition, we get

8x + 5y – 11 = 0

3x – 4y – 10 = 0

Writing the co-efficient in the following way, we get: 

Note: The above presentation is not compulsory for solving. 

By cross-multiplication method: 

x/(5) (-10) – (-4) (-11) = y/(-11) (3) – (-10) (8) = 1/(8) (-4) – (3) (5) 

or, x/-50 – 44 = y/-33 + 80 = 1/-32 – 15

or, x/-94 = y/47 = 1/-47

or, x/-2 = y/1 = 1/-1 [multiplying by 47] 

or, x = -2/-1 = 2 and y = 1/-1 = -1

Therefore, required solution is x = 2, y = -1



2. Find the value of x and y by using the using cross-multiplication method: 

3x + 4y – 17 = 0

4x – 3y – 6 = 0

Solution:

Two given equations are: 

3x + 4y – 17 = 0

4x – 3y – 6 = 0

By cross-multiplication, we get: 

x/(4) (-6) – (-3) (-17) = y/(-17) (4) – (-6) (3) = 1/(3) (-3) – (4) (4) 

or, x/(-24 – 51) = y/(-68 + 18) = 1/(-9 – 16) 

or, x/-75 = y/-50 = 1/-25

or, x/3 = y/2 = 1 (multiplying by -25) 

or, x = 3, y = 2

Therefore, required solution: x = 3, y = 2. 



3. Solve the system of linear equations: 

ax + by – c² = 0

a²x + b²y – c² = 0


Solution:

x/(-b + b²) = y/(- a² + a) = c²/(ab² - a²b) 

or, x/-b(1 - b) = y/- a(a - 1) = c²/-ab(a - b) 

or, x/b(1 - b) = y/a(a - 1) = c²/ab(a - b) 

or, x = bc²(1 – b)/ab(a – b) = c²(1 – b)/a(a – b) and y = c²a(a – 1)/ab(a – b) = c²(a – 1)/b(a – b) 

Hence the required solution is: 

x = c²(1 – b)/a(a – b) 

y = c²a(a – 1)/b(a – b)

Answer 2

taking LCM,

x^2-2x+3x-6+x^2-3x+2x-6/x^2-2x-3x+6 = 2

solving,

2x^2 -2x -12/x^2-5x+6=2
x^2-x-6 = x^2 -5x +6
so like terms cancel out then,

4x = 12

x = 3

hope this helps you out!