<p>Can someone explain how to do this question? Thanks</p>
<p>bumppppppppppppp</p>
<p>Well here what i got to…
z=x-y+4 Change this to y=-z+x+4
z=y-w-3 put it into this z=(-z+x+4)-w-3 =2z=x-w+1
z=w-x+5 now put this into z----------2(w-x+5)=x-w+1
3w+9=3x ---- x=w+3
now put this into the last equation----z=w-(w+3)+5 and z will be 2</p>
<p>Add the three equations up and get 3z = 6, z=2</p>
<p>I knew my way was too much for sat,however when i looked at the problem, that didn’t come to my mind lol. Sorry!</p>
<p>ok thanks everybody</p>
<p>can u explain why ur allowed (mathematically) to just add them up to solve?</p>
<p>I think it’s called linear combination. Basically, an equation is two expressions combined using a sign (<,>,=). Since you are using an = sign, both sides have the same value. If you add any polynomial to both sides of an expression, both sides are still equal. Since you are respective sides of an equation, you are essentially adding the same value to both sides, so the equation is still equal.
If you want to show that in a much more drawn out way, here it is:</p>
<p>z=x-y+4
(z)+z=(x-y+4)+z --add ‘z’ to both sides
z=y-w-3
(z)+z=(x-y+4)+(y-w-3) – substitute 2nd equation for ‘z’
[(z)+z]+z=[(x-y+4)+(y-w-3)]+z – add ‘z’ to both sides
z=w-x+5
[(z)+z]+z=[(x-y+4)+(y-w-3)]+(w-x+5) – substitute 3rd equation for ‘z’
3z=6 – combine like terms
z=2 – divide each side by 2</p>
<p>Wow, that was ridiculous. Maybe someone else can explain it better…</p>