Question

Solve the following pairs of linear equations by cross-multiplication method: 7x - 4y= 49; 5x - 6y = 57.
By elimation method ​

Answer 1

Answer:

Step-by-step explanation:

By Cross-multiplication method,

given equations,

7x-4y=49---(1)

5x-6y=57----(2)

by comparing both the equations with 'ax²+by+c=0' form we get

7x-4y-49=0

a₁=7 , b₁=-4 , c₁=-49   ------(3)

5x-6y-57=0                                                                                                                      

a₂=5 , b₂=-6 , c₂=-57  -----(4)

by cross multiplication method,

$\frac{x}{b_{1}c_{2}-b_{2}c_{1}}$  = $\frac{y}{c_{1}a_{2}-c_{2}a_{1}}$  = $\frac{1}{a_{1}b_{2}-a_{2}b_{1}}$        

substituting the values  from equations (3) and (4),

$\frac{x}{228-294}$  = $\frac{y}{-245+399}$ = $\frac{1}{-42+20}$

             

$\frac{x}{-66}$ = $\frac{y}{154}$ = $\frac{1}{-22}$ aaaaaaa

(I)     (II)    (III)

comparing (I) and (III)

$\frac{x}{-66}$=$\frac{1}{-22}$

$\frac{x}{1}$ = $\frac{1*-66}{-22}$

x=3

comparing (II) and (III),

$\frac{y}{154}$ = $\frac{1}{-22}$

$\frac{y}{1}$ = $\frac{1*154}{-22}$

y=-7

By Elimination method,

7x-4y=49 ---(1)

5x-6y=57 ---(2)

multiplying eqn (1) and (2) with 3 and 2 respectively

21x-12y=147---(5)

10x-12y=114----(6)

subtracting eqn (5) and (6) we get,

   21x-12y=147

  -10x+12y=-114

    11x = 33

x=33/11

x=3

substituting the value of x in eqn (2),

5x-6y=57

5(3)-6y=57

$\frac{y}{1} = \frac{57-15}{-6}$

y=$\frac{42}{-6}$

y=-7

 hence the given pairs of equation are solved using cross-multiplication and elimination respectively

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