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Method of Substitution to Find the Solution of a Pair of Linear Equations

6 x+5 y=82 x+...

Question

6x +5y= 8

2x +3y=5

solve this linear equation by 20 three methods substitution ,elimination and cross multiplication

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Solution

(We have no permission to solve it on a paper . Dont worry, i will explain every thing simple as possible)

The given question is to solve equations 6x +5y= 8 and 2x +3y=5 .

(i) By substitution:
$F r o m space e q u a t i o n space 2 comma 2 x plus 3 y equals 5 S o space 2 x equals 5 minus 3 y O r space x equals fraction numerator 5 minus 3 y over denominator 2 end fraction$
Now put this value of x in equation(1)
$6 x plus 5 y equals 8 rightwards double arrow 6 cross times open parentheses fraction numerator 5 minus 3 y over denominator 2 end fraction close parentheses plus 5 y equals 8 space rightwards double arrow 3 cross times open parentheses 5 minus 3 y close parentheses plus 5 y equals 8 space left parenthesis because space 6 over 2 equals 3 right parenthesis rightwards double arrow 3 cross times 5 minus 3 cross times 3 y plus 5 y equals 8 rightwards double arrow 15 minus 9 y plus 5 y equals 8 rightwards double arrow space 15 minus 4 y equals 8 rightwards double arrow 15 minus 8 equals 4 y O r space 7 equals 4 y S o space v a l u e space o f space y equals 7 over 4$
Similiarly , from equatio(2)
$2 x plus 3 y equals 5 O r space 3 y equals 5 minus 2 x S o space y equals fraction numerator 5 minus 2 x over denominator 3 end fraction$
Now put this value of y in equation(1)
$6 x plus 5 y equals 8 rightwards double arrow 6 x plus 5 cross times open parentheses fraction numerator 5 minus 2 x over denominator 3 end fraction close parentheses equals 8 rightwards double arrow 6 x plus fraction numerator 5 cross times 5 minus 2 x cross times 5 over denominator 3 end fraction equals 8 rightwards double arrow fraction numerator 6 x cross times 3 plus 25 minus 10 x over denominator 3 end fraction equals 8 rightwards double arrow fraction numerator 18 x plus 25 minus 10 x over denominator 3 end fraction equals 8 rightwards double arrow fraction numerator 8 x plus 25 over denominator 3 end fraction equals 8 rightwards double arrow 8 x plus 25 equals 24 rightwards double arrow 8 x equals 24 minus 25 rightwards double arrow 8 x equals negative 1 O r space x equals negative 1 over 8$
(ii) By elimination

Now multiply Equation(2) by 3
$3 cross times open parentheses 2 x plus 3 y equals 5 close parentheses rightwards double arrow 3 cross times 2 x plus 3 cross times 3 y equals 3 cross times 5 rightwards double arrow 6 x plus 9 y equals 15 L e t space 6 x plus 9 y equals 15 space............ E q u a t i o n left parenthesis 3 right parenthesis N o w space s u b s t r a c t space E q u a t i o n left parenthesis 1 right parenthesis space f r o m space E q u a t i o n left parenthesis 3 right parenthesis W e space g e t space comma fraction numerator 6 x plus 9 y equals 15 space open parentheses minus close parentheses 6 x plus 5 y equals 8 over denominator 0 x plus 4 y equals 7 end fraction O r space y equals 7 over 4$
Similiarly, for finding the value of x,
$m u t l i p l y space e q u a t i o n left parenthesis 1 right parenthesis space b y space 3 comma space w e space g e t 18 x plus 15 y equals 24 L e t space 18 x plus 15 y equals 24 space........ E q u a t i o n left parenthesis 4 right parenthesis N o w space m u l t i p l y space e q u a t i o n left parenthesis 2 right parenthesis space b y space 5 comma space w e space g e t 10 x plus 15 y equals 25 L e t space 10 x plus 15 y equals 25.......... E q u a t i o n left parenthesis 5 right parenthesis N o w space s u b s t r a c t space e q u a t i o n left parenthesis 5 right parenthesis space f r o m space E q u a t i o n left parenthesis 4 right parenthesis space W e space g e t comma fraction numerator 18 x plus 15 y equals 24 space open parentheses minus close parentheses 10 x plus 15 y equals 25 over denominator 8 x plus 0 y equals negative 1 end fraction T h a t space i s comma space 8 x equals negative 1 O r space x equals negative 1 over 8$
(iii) Cross multiplication method

$fraction numerator x over denominator open parentheses 5 cross times negative 5 close parentheses minus open parentheses 3 cross times negative 8 close parentheses end fraction equals fraction numerator y over denominator open parentheses negative 8 cross times 2 close parentheses begin display style minus end style begin display style open parentheses negative 5 cross times 6 close parentheses end style end fraction equals fraction numerator 1 over denominator open parentheses 6 cross times 3 close parentheses begin display style minus end style begin display style open parentheses 2 cross times 5 close parentheses end style end fraction rightwards double arrow fraction numerator x over denominator negative 25 minus open parentheses negative 24 close parentheses end fraction equals fraction numerator y over denominator negative 16 minus open parentheses negative 30 close parentheses end fraction equals fraction numerator 1 over denominator 18 minus 10 end fraction rightwards double arrow fraction numerator x over denominator negative 1 end fraction equals y over 14 equals 1 over 8 N o w space fraction numerator x over denominator negative 1 end fraction equals 1 over 8 space rightwards double arrow x equals negative 1 cross times 1 over 8 equals fraction numerator negative 1 over denominator 8 end fraction A n d space y over 14 equals 1 over 8 rightwards double arrow y equals 14 over 8 equals 7 over 4$

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