3x_5y=0 9x=2y+7

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3x_5y=0 9x=2y+7

Posted by Raman Rai 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

By Elimination method,
The given system of equations is :
3 x - 5 y - 4 = 0............(1)
9 x = 2 y + 7
9 x - 2 y - 7 = 0.............(2)
Multiplying equation (1) by 3, we get
9 x - 15 y - 12 = 0.............(3)
Subtracting equation (3) from equation (2) , we get
13 y + 5 = 0
{tex}\Rightarrow \quad 13 y = - 5 \Rightarrow y = \frac { - 5 } { 13 }{/tex}
Substituting this value of y in equation (1), we get
{tex}3 x - 5 \left( \frac { - 5 } { 13 } \right) - 4 = 0{/tex}
{tex}\Rightarrow \quad 3 x + \frac { 25 } { 13 } - 4 = 0 \Rightarrow 3 x - \frac { 27 } { 13 } = 0{/tex}
{tex}\Rightarrow \quad 3 x = \frac { 27 } { 13 } \Rightarrow x = \frac { 9 } { 13 }{/tex}
So, the solution of the given system of equation is
{tex}x = \frac { 9 } { 13 } , y = \frac { - 5 } { 13 }{/tex}
By Substitution method:
The given system of equation is:
3 x - 5 y - 4 = 0.............(1)
9 x = 2 y + 7...................(2)
From equation (2),
{tex}x = \frac { 2 y + 7 } { 9 }{/tex}..................(3)
Substituting this value of x in equation(1), we get
{tex}3 \left( \frac { 2 y + 7 } { 9 } \right) - 5 y - 4 = 0{/tex}
{tex}\Rightarrow \quad \frac { 2 y + 7 } { 3 } - 5 y - 4 = 0{/tex}
{tex}\Rightarrow \quad 2 y + 7 - 15 y - 12 = 0{/tex}
{tex}\Rightarrow{/tex} -13y - 5 = 0
{tex}\Rightarrow{/tex} 13y = -5
{tex}\Rightarrow \quad y = \frac { - 5 } { 13 }{/tex}
Substituting this value of y in equation(3), we get
{tex}x = \frac { 2 \left( \frac { - 5 } { 13 } \right) + 7 } { 9 } = \frac { - \frac { 10 } { 13 } + 7 } { 9 } = \frac { - 10 + 91 } { 117 } = \frac { 81 } { 117 } = \frac { 9 } { 13 }{/tex}

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