"Solve:[tex]2y = x + 2[/tex] and [tex]x - 3y = -5[/tex]?"...
Solve using substitution: [tex]2y=x+2[/tex] [tex]y=(x+2)/2[/tex] ---------------------------- [tex]x -3y = -5[/tex] is also [tex]x - 3((x+2)/2) = -5[/tex] Solve for x. [tex]x - 3((x+2)*1/2) = -5[/tex] Distribute the parenthesis. [tex]x - 3(x*1/2+2*1/2) = -5[/tex] Simplify. [tex]x - 3(1/2x+1) = -5[/tex] Distribute the parenthesis. [tex]x - (3*1/2x+3*1)= -5[/tex] Simplify [tex]x - (3/2x+3)= -5[/tex] Add -3 to both sides. [tex](x - (3/2x+3))+(-3)=(-5)+(-3)[/tex] Simplify. [tex]x - (3/2x)=-2[/tex] Find LCM. [tex]2/2x-(3/2x)=-2[/tex] [tex]-1/2x=-2[/tex] Multiply both sides by -2. [tex](-1/2x)*(-2)=(-2)*(-2)[/tex] Simplify. [tex]x=4[/tex]
Now Input this constant in place of the variable x. [tex]2y = x + 2[/tex] [tex]2y = 4+2[/tex] Add like terms. [tex]2y = 6[/tex] Divide both sides by 2. [tex](2y)/2 = (6)/2[/tex] Simplify. [tex]y=3[/tex]
