All triangles have the same value, and all circles have the same value. What is the sum of three circles?
Let the triangle = $x$ and the circle = $y$.
Thus, the two equations are
$3x + 2y = 21$ and $3y + 2x = 19$.
Since we want to find $y$, we multiply the second equation by $\frac{3}{2}$ to get $\frac{9y}{2} + 3x = \frac{57}{2}$.
Subtracting the second and first equations gives $\frac{5y}{2} = \frac{15}{2}$
Thus, $y = 3$, which is the value of the circle. Can you solve it from here?
This question may have been asked by a child who has never used algebra.
It is easier to use algebra for a person who knows how BUT here is how another way to do it.
the first one has 2 circles, 2 trangles and another triangle and it adds to 21
the second one has 2 circles, 2 trangles and another circle and it adds to 19
So the triangle must be bigger than the circle by 2. (because that is the only difference)
So if I replace each of the triangles in the second one with a circle plus 2 more I get
5 circles + 2+2 = 19
If 5 circles plus 4 more equals 19 then
5 circles must equal 15
so one circle must be 3 (because 3+3+3+3+3=15)