The solution set of x+y-1=0 and 3x+3y-2=0 is | vedclass

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(A) Given equations are:$x+y-1=0 \implies x+y=1$  (Equation $1$)$3x+3y-2=0 \implies 3(x+y)=2 \implies x+y=\frac{2}{3}$ (Equation $2$)Comparing the coe

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$\phi$

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(A) Given equations are:$x+y-1=0 \implies x+y=1$  (Equation $1$)$3x+3y-2=0 \implies 3(x+y)=2 \implies x+y=\frac{2}{3}$ (Equation $2$)Comparing the coefficients of the system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$\frac{a_1}{a_2} = \frac{1}{3}$,$\frac{b_1}{b_2} = \frac{1}{3}$,and $\frac{c_1}{c_2} = \frac{-1}{-2} = \frac{1}{2}$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$,the lines are parallel and do not intersect.Therefore,there is no solution to this system of equations.The solution set is $\phi$ (the empty set).

The solution set of $x+y-1=0$ and $3x+3y-2=0$ is ...........

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