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You can set this problem up like this: 48/4 = y/11 Cross multiply 48*11 = 4y 528 = 4y 528 divided by 4 = 132 So when w equals 11 y equals 132
14y minus 6y means "14 times a number, minus 6 times the same number". If you have 14 things and you take away 6 things, you are left with 8 things. So, 8 times a number equals 48. 8 times what number gets you 48? Since you are already in Algebra I, you know 8 times 6 gets you 48, so the number (y) is 6. The formal Algebra I approach is to combine like terms, then divide. 14y - 6y = 48 (Subtract 6y from 14 y. (14y - 6y = 8y, and the y does not go away) 8y = 48 (Now divide both sides by 8 to get the y all by itself) 8/8y = 48/8 (8/8 is 1, and 48/8 is 6, so 1y = 6. You can lose the 1 because one thing is the thing itself.) y = 6 Does that help? There are many ways to solve this problem, but Algebra I is teaching the proper steps for solving one, two and three step equations. Stick with it.
do you mean 16 = y / 3 ?? if so, y = 48.
y = arcsin( cos 48 ); arcsin may be seen as sin-1 on your calculator.
y = 15 (by transitivity)