Maths Olympiad Prep

Library / /342 of 860

Algebra Difficulty 5.1 AIME, harder Find the answer

The real numbers x,y,z,wx, y, z, w satisfy 2x+y+z+w=1x+3y+z+w=2x+y+4z+w=3x+y+z+5w=25\begin{aligned} & 2 x+y+z+w=1 \\ & x+3 y+z+w=2 \\ & x+y+4 z+w=3 \\ & x+y+z+5 w=25 \end{aligned} Find the value of ww.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Multiplying the four equations by 12,6,4,312,6,4,3 respectively, we get 24x+12y+12z+12w=126x+18y+6z+6w=124x+4y+16z+4w=123x+3y+3z+15w=75\begin{aligned} 24 x+12 y+12 z+12 w & =12 \\ 6 x+18 y+6 z+6 w & =12 \\ 4 x+4 y+16 z+4 w & =12 \\ 3 x+3 y+3 z+15 w & =75 \end{aligned} Adding these yields 37x+37y+37z+37w=11137 x+37 y+37 z+37 w=111, or x+y+z+w=3x+y+z+w=3. Subtract this from the fourth given equation to obtain 4w=224 w=22, or w=11/2w=11 / 2.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.