How do you solve the system $\frac{1}{3} x - y = 3$ $1/3x-y=3$ and $2 x + y = 25$ $2x+y=25$ using substitution?

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Answer 1 · 2017-10-02T05:43:38

$x = 12$ $x = 12$ and $y = 1$ $y=1$

Given equations are:

(1) ------ $\frac{1}{3} x - y = 3$ $1/3x-y=3$

(2)------ $2 x + y = 25$ $2x+y=25$

Multiply (1) by 3 to eliminate the fractional part,

(1) ------- $x - 3 y = 9$ $x- 3y = 9$

$\Rightarrow x = 9 + 3 y$ $=> x= 9+ 3y$ -------- let this be equation (3)

Now substitute this value of $x$ $x$ from (3) in equation (2),

(2) --------- $2 (9 + 3 y) + y = 25$ $2(9+3y) + y = 25$

$\Rightarrow 18 + 6 y + y = 25$ $=> 18 + 6y +y = 25$

$\Rightarrow 7 y = 25 - 18$ $=> 7y = 25- 18$

$\Rightarrow y = \frac{7}{7}$ $=> y = 7/7$

Therefore,

$y = 1$ $y=1$

Substituting this value of $y$ $y$ in equation (3),

$x = 9 + 3 \times 1$ $x= 9 + 3\times 1$

$x = 9 + 3$ $x = 9+3$

Therefore,
$x = 12$ $x = 12$
So we have the values of $x$ $x$ and $y$ $y$ as,

$x = 12$ $x = 12$ and $y = 1$ $y=1$

How do you solve the system 1/3x-y=313x−y=31/3x-y=3 and 2x+y=252x+y=252x+y=25 using substitution?