Ответ:

[tex](x;y)=\bigg(-\frac{33}{38} ;1\frac{35}{38} \bigg)[/tex]

Объяснение:

[tex]\displaystyle\left \{ {{7x+8=y,} \atop {3x+5y=7;}} \right. \Leftrightarrow \left \{ {{35x+40=5y,} \atop {3x+5y=7;}} \right. \\35x+3x+40+5y=5y+7;\\38x+40+\not5y-\not5y=7;\\38x=-33;\\x=-\frac{33}{38} .\\3x+5y=7;\\5y=7-3x;\\y=\frac{7-3x}{5} =\frac{7-3*\bigg(-\frac{33}{38}\bigg) }{5} =\frac{7^{(38}+\frac{99}{38} }{5} =\frac{\frac{266+99}{38} }{5} =\frac{\not365}{38} *\frac{1}{\not5} =\frac{73}{38} =1\frac{35}{38}[/tex]