+0  
 

Best Answer 

 #1
avatar+26400 
+30

(y^2+5y-7)-(y^2-3y-4)

 

\(\begin{array}{rcll} (y^2+5y-7)-(y^2-3y-4) &=&\\ &=& y^2+5y-7-(y^2-3y-4)\\ &=& y^2+5y-7-y^2+3y+4\\ &=& y^2-y^2+5y+3y-7+4 \qquad & | \qquad y^2-y^2 = 0\\ &=& 5y+3y-7+4 \qquad & | \qquad 5y+3y= 8y\\ &=& 8y-7+4 \qquad & | \qquad-7+4 =-3\\ &=& 8y-3 \\ \mathbf{(y^2+5y-7)-(y^2-3y-4)} &\mathbf{=}& \mathbf{8y-3 } \end{array}\)

 

laugh

 Oct 29, 2015
 #1
avatar+26400 
+30
Best Answer

(y^2+5y-7)-(y^2-3y-4)

 

\(\begin{array}{rcll} (y^2+5y-7)-(y^2-3y-4) &=&\\ &=& y^2+5y-7-(y^2-3y-4)\\ &=& y^2+5y-7-y^2+3y+4\\ &=& y^2-y^2+5y+3y-7+4 \qquad & | \qquad y^2-y^2 = 0\\ &=& 5y+3y-7+4 \qquad & | \qquad 5y+3y= 8y\\ &=& 8y-7+4 \qquad & | \qquad-7+4 =-3\\ &=& 8y-3 \\ \mathbf{(y^2+5y-7)-(y^2-3y-4)} &\mathbf{=}& \mathbf{8y-3 } \end{array}\)

 

laugh

heureka Oct 29, 2015

1 Online Users