We see the two last rows at the bottom, which clearly contain the same number of circles, except for the triangle and the square. We also understand the value of the two rows; 18 and 20. Therefore, because the difference between between a triangle and a square is 2, the square is 2 greater than the triangle.
We can suppose that 3circles+1square=20. We then substitute the shapes by variables; the circle=c, and the square=s. Thus, the equation is now 3c+s=20. For the second equation, the triangle=t. Therefore, the equation would be: 3c+t=18. In the first equation, we know that s=t+2. Therefore, we can substitute the value of s in terms of t and a constant into the equation. 3c+s=20 becomes: 3c+t+2=20. When we simplify the equation, we want to have like terms on the right side of the equal sign, so t=18-3c.
Now, by using the second column from the left side of the table, we can find the solution. We know that the s=t+2. In this column, however, there are 3 squares and 1 triangle. Therefore, we multiply 3(t+2) and get 3t+6 as the result. We now know that using the total value of the shapes, 30=3t+6+t, we then simplify the equation. 4t+6=30. Subtract 6 from the left and right side, we get 4t+6-6=30-6 OR 4t=24. Finally, to isolate t, we divide 4t by 4 and 24 by 4. Thus, the triangle=6.
Now, to find the value of the circle, we use our previous equation, t=18-3c, and substitute the the t by 6. Just by simplifying as we were doing, 6=18-3c, and the circle equals 4.
Because we know that the square=t+2, the square=6+2. Therefore, the square=8. Finally, for the hexagon, we can find the 2nd row form the top to have 2 hexagons and 2 squares. If the hexagon=h, 2h+2s=30. 2h+2(8)=30. 2h-16=30. Thus, the hexagon=7.
For the first column from the left, we have 1triangle+1hexagon+2circles=?
t=6, h=7, c=4. 6+7+2(4)=? 6+7+8=? Thus, the solution to this problem is: ?=21