##x=-5##

When we have a fraction equal to another fraction we can use the method of ##color(blue)”cross-multiplication”## to solve.

That is ##(color(red)(x+2))/(color(blue)(x-1))=color(blue)(1)/color(red)(2)##

Now multiply the terms on either end of an ‘imaginary’ cross (X). That is multiply the ##color(red)(“red”)## values together and the ##color(blue)(“blue”)## values together and equate them.

##rArrcolor(red)(2(x+2))=color(blue)(1(x-1))##

distribute the brackets.

##rArr2x+4=x-1##

We now want to have the x terms on the left of the equation and numeric values on the right.

subtract x from both sides.

##2x-x+4=cancel(x)cancel(-x)-1##

##rArrx+4=-1##

subtract 4 from both sides.

##xcancel(+4)cancel(-4)=-1-4##

##rArrx=-5″ is the solution”##