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zysi [14]
3 years ago
8

How is this solved. Ive tried different ways and cant get it

Mathematics
1 answer:
lukranit [14]3 years ago
3 0
It's kinda like using variables, but shapes in their place. Let's set the circle to x, the triangle to y, and the square to z. Now the equations say x+(y+z)=13, (y+z)=8, and x+y=7. First thing we can do it plug 8 into the first equation for y+z, getting x+8=13. Subtract 8 to both sides and x (or circle) equals 5. Plug 5 into the last equation, x+y=7 to solve for y, getting 5+y=7. Subtract 5 to both sides and y (or triangle) equals 2. Finally, plug two in for y in y+z=8, getting 2+z=8. Subtract 2 from both sides and z (or square) equals 6. Hope I've explained it clearly!!
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The answer to your question is:

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