Question

Вот, задание ниже, пж скорее...

Answer 1

a) одз:
$4y+10 neq 0 \4y neq -10 \y neq -2,5 \3y-4 neq 0 \3y neq 4 \y neq frac{4}{3}$
решаем:
$left { {{3(3x-1)=4y+10} atop {7(5x+1)=-6(3y-4)}} right. \ left { {{9x-3=4y+10} atop {35x+7=-18y+24}} right. \ left { {{9x-4y=13} atop {35x+18y=17}} right. \y= frac{9x-13}{4} \35x+ frac{18*(9x-13)}{4} =17 \70x+9(9x-13)=34 \70x+81x-117=34 \151x=151 \x=1 \y= frac{9-13}{4} =-1$
Ответ: (1;-1)
b) одз:
$b neq - frac{1}{3}$
решаем:
$9b+3= 3(4a-10) \ frac{5a-7}{3(4a-10)} =- frac{1}{2} \-10a+14=12a-30 \-22a=-14-30 \-22a=-44 \a=2 \9b+3=3(8-10) \9b=-6-3 \9b=-9 \b=-1$
Ответ: a=2; b=-1