Question

solve the following pair of linear equationby substitution and elimination method. 2x+3y=5 and 3x-2y=1​

Answer 1

Step-by-step explanation:

given that

eq.1.= 2x+3y=5

eq. 2.=3x-2y=1

we can write the eq.2. as

x= (1+2y)/3......[eq.3.]

now, by substituting the value of x from eq. 3 in eq. 1. we get

2×(1+2y)/3+3y=5

[2+4y+9y]/3=5

2+13y=15

13y=15-2

13y=13

y=1.........(¥)

now putting the value of y from(¥) in eq. 3. we get the value of x

x=(1+(2×1)/3

=3/3=1

x=1

Answer 2

Answer:

The solution of the given simultaneous equations is

( x, y ) = ( 1, 1 ).

Step-by-step-explanation:

We have given two simultaneous equations.

We have to find their solutions by substitution and elimination methods.

1. Substitution method:

$\sf\:2x\:+\:3y\:=\:5\\\\\implies\sf\:2x\:=\:5\:-\:3y\\\\\implies\sf\:x\:=\:\dfrac{5\:-\:3y}{2}\\\\\implies\sf\:x\:=\:\dfrac{-\:3y\:+\:5}{2}\:\:\:-\:-\:(\:1\:)\\\\\sf\:3x\:-\:2y\:=\:1\\\\\implies\sf\:3\:\times\:(\:\dfrac{\:-\:3y\:+\:5}{2}\:)\:-\:2y\:=\:1\:\:\:-\:-\:[\:From\:(\:1\:)\:]\\\\\implies\sf\:\dfrac{-\:9y\:+\:15}{2}\:-\:2y\:=\:1\\\\\implies\sf\:-\:9y\:+\:15\:-\:4y\:=\:2\:\:\:-\:-\:-\:[\:Multiplying\:by\:2\:]\\\\\implies\sf\:-\:13y\:=\:2\:-\:15\\\\\implies\sf\:\cancel{-}\:13y\:=\:\cancel{-}\:13\\\\\implies\sf\:y\:=\:\cancel{\frac{13}{13}}\\\\\implies\boxed{\red{\sf\:y\:=\:1}}$

By substituting y = 1 in equation ( 1 ), we get,

$\sf\:x\:=\:\dfrac{-\:3\:y\:+\:5}{2}\:\:\:-\:-\:(\:1\:)\\\\\implies\sf\:x\:=\:\dfrac{-\:3\:\times\:1\:+\:5}{2}\\\\\implies\sf\:x\:=\:\dfrac{\:-\:3\:+\:5}{2}\\\\\implies\sf\:x\:=\:\cancel{\frac{2}{2}}\\\\\implies\boxed{\red{\sf\:x\:=\:1}}$

2. Elimination method:

$\sf\:2x\:+\:3y\:=\:5\:\:\:-\:-\:(\:1\:)\\\\\sf\:3x\:-\:2y\:=\:1\:\:\:-\:-\:(\:2\:)$

By multiplying equation ( 1 ) by 2 and equation ( 2 ) by 3 and adding both equation, we get,

$\sf\:4x\:+\:\cancel{6y} \:=\:10\:\:\:-\:-\:(\:3\:)\\\\\sf\:\underline{\sf\:9x\:-\:\cancel{6y}\:=\:3}\sf\:\:\:-\:-\:(\:4\:)\\\\\implies\sf\:13x\:=\:13\\\\\implies\sf\:x\:=\:\cancel{\frac{13}{13}}\\\\\implies\boxed{\red{\sf\:x\:=\:1}}$

Now, by substituting x = 1 in equation ( 1 ), we get,

$\sf\:2x\:+\:3y\:=\:5\:\:\:-\:-\:(\:1\:)\\\\\implies\sf\:2\:\times\:1\:+\:3y\:=\:5\\\\\implies\sf\:2\:+\:3y\:=\:5\\\\\implies\sf\:3y\:=\:5\:-\:2\\\\\implies\sf\:3y\:=\:3\\\\\implies\sf\:y\:=\:\cancel{\frac{3}{3}}\\\\\implies\boxed{\red{\sf\:y\:=\:1}}$