Решите уравнение

без одз! ​

Объяснение:

<var>{

15x−3y=−3

5x−2y=1

</var>

< var > \left \{ {{5x-2y=1} \atop {3y=15x+3}} \right. < /var ><var>{

3y=15x+3

5x−2y=1

</var>

< var > \left \{ {{5x-2y=1} \atop {y=\frac{15x+3}{3}}} \right. < /var ><var>{

y=

3

15x+3

5x−2y=1

</var>

< var > \left \{ {{5x-2y=1} \atop {y=\frac{15x}{3}+\frac{3}{3}}} \right. < /var ><var>{

y=

3

15x

+

3

3

5x−2y=1

</var>

< var > \left \{ {{5x-2y=1} \atop {y=5x+1}} \right. < /var ><var>{

y=5x+1

5x−2y=1

</var>

< var > \left \{ {{y=5x+1} \atop {5x-2\cdot(5x+1)=1}} \right. < /var ><var>{

5x−2⋅(5x+1)=1

y=5x+1

</var>

< var > 5x-2\cdot(5x+1)=1 < /var ><var>5x−2⋅(5x+1)=1</var>

< var > 5x-10x-2=1 < /var ><var>5x−10x−2=1</var>

< var > -5x-2=1 < /var ><var>−5x−2=1</var>

< var > 5x=-2-1 < /var ><var>5x=−2−1</var>

< var > 5x=-3 < /var ><var>5x=−3</var>

< var > x=-3:5 < /var ><var>x=−3:5</var>

< var > x=-\frac{3}{5}=-0,6 < /var ><var>x=−

5

3

=−0,6</var>

< var > y=5\cdot(-0,6)+1=-3+1=-2 < /var ><var>y=5⋅(−0,6)+1=−3+1=−2</var>

< var > \left \{ {{x=-0,6} \atop {y=-2}} \right. < /var ><var>{

y=−2

x=−0,6

< var > \left \{ {{5\cdot(-0,6)-2\cdot(-2)=-3+4=1} \atop {15\cdot(-0,6)-3\cdot(-2)=-9+6=-3}} \right. < /var ><var>{

15⋅(−0,6)−3⋅(−2)=−9+6=−3

5⋅(−0,6)−2⋅(−2)=−3+4=1

</var>

прости если не правильно:)