How do you solve for x in #5x + 7x = 72#?

Answer 1

It's crucial, in my opinion, to be able to solve problems as well as comprehend why a specific approach might be taken.

The first step to solve the above equation is to simplify it using one of the laws of arithmetic - the distributive law that states that #a*c+b*c=(a+b)*c#
Using this law, we can say that #5*x+7*x=(5+7)*x# Since #5+7=12#, we can rewrite our equation in a form #12*x=72#

The equation will then be transformed using a different law, which says that two equal numerical values can be divided by the same non-zero number, and the resulting values will also be equal.

Let's divide both side of the equation above by #12#, the result is #(12*x)/12=72/12#
Performing obvious operations, we get #x=6# This is the solution since we have determined the value of an unknown variable #x#.
I would strongly recommend to check the answer obtained, no matter how simple the transformation steps were. In this case we can substitute the value #x=6# into original equation and check if it's true. Indeed, #5*6+7*6=30+42=72#, which proves that the solution we obtained #x=6# is correct.
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Answer 2

To solve for ( x ) in the equation ( 5x + 7x = 72 ), you first combine like terms on the left side of the equation to get ( 12x = 72 ). Then, divide both sides by 12 to isolate ( x ), yielding ( x = 6 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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