By converting the shapes to variables, I considered the triangle X, the square Y, and the circle Z.
The top row would equate to this: 2x+2y=28
The third row would equate to this: x+3z=18
The last row would equate to this: y+3z=20
Solving for X, the second equation could be arranged as so:
x=18-3z
Solving for Y, the third equation could be arranged as so:
y=20-3z
We now have an X and Y value which we could substitute into the first equation:
2(18-3z)+2(20-3z)=28
36-6z+40-6z=28
-6z+76=28
-6z=-48
-> z=4
-> Circle = 4
We now have an actual Z value, and by plugging this value into the second and third equations, we could find X and Y, respectively:
x+3(4)=18
x+12=18
-> x=6
-> Triangle = 6
y+3(4)=20
y=12=20
-> y=8
-> Square = 8
We now have a real value for the circle, triangle, and square.
To find the value of the hexagon, we can create an equation for the second row:
2y+2h=30
2(8)+2h=30
16+2h=30
2h=14
-> h=7
-> Hexagon = 7
We now have values for all the shapes.
In order to find the value of the first column, we could easily set up an equation:
1(6)+1(7)+2(4)=x ->[1 triangle, 1 hexagon, 2 circles]
6+7+8=x
-> x=21
In order to check your work, you could replace any shape with its value, and then compare your calculated value with the actual row/column value.