Question

Solve each of the following systems of equations by the method of cross-multiplication:
2x + y = 35
3x + 4y = 65

Answer 1

Concept :

CROSS - MULTIPLICATION METHOD:

The general form of a pair of linear equations

a1x + b1y + c1 = 0 , &        a2x + b2y + c2 = 0.

When a1 / a2 ≠ b1 / b2, the pair of linear equations will have a unique solution.

To solve this pair of equations for x and y using cross-multiplication, we’ll arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2  as shown below

⇒ x =  b1 c2 - b2 c1  / a1 b2 - a2 b1    

⇒ y =  c1 a2 - c2 a1  / a1 b2 - a2 b1

The above equation is generally written as :

x/ b1 c2 - b2 c1 =  y/ c1 a2 - c2 a1 = 1/a1 b2 - a2 b1

x                       y                        1

-----------   =   -----------------     =    ---------

b1      c1        c1           a1           a1        b1

b2      c2       c2           a2           a2       b2

Given :

2x + y = 35

3x + 4y = 65

 

Solution :  

We have ,  

2x + y - 35 = 0

3x + 4y - 65 = 0

Here a1 = 2, b1 = 1, c1 = - 35

a2 = 3, b2 = 4, c2 = - 65

x                       y                        1

-----------   =   -----------------     =    ---------

1      -35         - 35         2           2       1

4    -65           - 65       3            3     4

x/( 1 × - 65) - (4 × - 35) = y/(- 35 × 3) - (- 65 × 2) = 1 /(2 × 4) - (3 × 1)

x/- 65 + 140 = y/- 105 + 130 = 1/ 8 - 3

x/75 = y/25 = 1/5

Now, x/75  = 1/5

x =  75/5

x = 15

And  

y/25 = 1/5

y = 25/5

 y = 5

Hence the value of given systems of equations is x = 15 and y = 5.

Hope this answer will help you…

 

Some more questions from this chapter :  

Solve each of the following systems of equations by the method of cross-multiplication :

x + 2y + 1 = 0

2x - 3y - 12 = 0

https://brainly.in/question/15918701

 

Solve each of the following systems of equations by the method of cross-multiplication :

3x + 2y + 25 = 0

2x + y + 10 = 0

https://brainly.in/question/15918700

Answer 2

$\huge\bold\red{Answer:-}$

The Given system of equation may be written as

$2x + y - 35 = 0$

$3x + 4y - 65 = 0$

$\huge\bold\red{Here,}$

$a1 = 2,b1 = 6,c1 = 35$

$a2 = 3,b2 = 4 \: and \: c2 = 65$

By crpss multiplication,

$\huge\bold\red{We \: have,}$

$= > \frac{x}{1 + ( - 65) - ( - 35) \times 4} = \frac{ - y}{2x( - 65) - ( - 35) \times 3}$

$= \frac{1}{2x4 - 1 \times 3}$

$= > \frac{x}{ - 65 + 140} = \frac{x}{ - 130 - 150} = \frac{1}{8 - 3}$

$= > \frac{x}{75} = \frac{ - y}{ - 25} = \frac{1}{5}$

$\frac{x}{75} = \frac{y}{25} = \frac{1}{5}$

$\huge\bold\red{Now,}$

$\frac{y}{25} = \frac{1}{5}$

$= > y = \frac{25}{5} = 5$

$\huge\bold\red{Hence,}$

X = 15, y = 5 the solution of the given system of equation.