Answer
$(1, 3)$
Work Step by Step
Given system is-
$26x -10 y$ = $-4$ __ eq.1
$-0.6x +1.2 y$ = $3$ __ eq.2
Multiplying eq.1 by '12' and eq.2 by '100'-
$312x -120 y$ = $-48$ __ eq.3
$-60x +120 y$ = $300$ __ eq.4
Adding eq.3 and eq.4 -
$312x -120 y-60x +120 y$ = $-48+300$
i.e. $312x -60x$ = $252$
i.e. $252x$ = $252$
i.e. $x$ = $1$
Substituting for $x$ in eq.1
$26(1) -10 y$ = $-4$
i.e. $26 -10y$ = $-4$
i.e. $10y$ = $26+4$
i.e. $10y$ = $30$
i.e. $y$ = $3$
Thus $(1, 3)$ is the solution of given system.
