How do you solve the system of equations $4 x + 3 y = 13$ $4x + 3y = 13$ and $6 x + 2 y = 17$ $6x + 2y = 17$ ?

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Answer 1 · 2017-08-13T13:24:06

see below

$4 x + 3 y = 13$ $4x+3y=13$ , $6 x + 2 y = 17$ $6x+2y=17$

Multiply equation $(2)$ $(2)$ by $\frac{3}{2}$ $3/2$ ,

$\Rightarrow 9 x + 3 y = \frac{51}{2}$ $=>9x+3y=51/2$

Subtract equation $(1)$ $(1)$ from this obtained equation,

$\Rightarrow 5 x = \frac{25}{2}$ $=>5x=25/2$

$\Rightarrow x = \frac{5}{2}$ $=>x=5/2$

Now anyone can easily solve for $y$ $y$ ,

$y = \frac{17 - 6 x}{2} = \frac{17 - 15}{2} = 1$ $y=(17-6x)/2=(17-15)/2=1$

So,

$x = \frac{5}{2}$ $x=5/2$ , $y = 1$ $y=1$

How do you solve the system of equations 4x+3y=134x + 3y = 13 and 6x+2y=176x + 2y = 17?