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Two identical integers whose sum is 66 are 33 and 33. Add 1 and subtract 1. Now you have 32 and 34. Voilà : two consecutive even integers whose sum is 66.
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My answer here is the same as the answer above, I'm just going to explain in detail how to actually solve a problem like this. Two consecutive even integers would be like 2 and 4 or 632 and 634. See how the numbers come right after each other, that would be a consecutive integer. Since you wanted consecutive even integers, they are 2 even numbers that come after each other. Now, to solve your problem, the 'formula' for finding 2 consecutive even (and odd, plain consecutive integers would be different) integers for the sum of 66 would be:
x + x+2 = 66. First, you add like terms. What I mean, is that you add all your x values and all your normal numbers, which gets us:
2x + 2 = 66 Now, you have a normal 2-step equation. To solve a 2-step equation, first you tik-tok. This means if you have any subtraction, you change it so that you add negatives instead(none here). For example 6x-3 changes into 6x + -3. Then you add like terms, which I already explained somewhere above (none here). Then, you add the opposite of your lonesome number(number without a variable) to both sides of the equation. So for example here, you add -2 to both sides of the equation so that you can get rid of that lonesome number, like here:
2x + (2 + -2)=66 + -2 -----> 2x = 64 ( no need to put + 0) Now all you have left to do is divide both sides of the equation by the coefficient of the x term, here being 2. LOOK TO SEE IF IT's POSITIVE OR NEGATIVE!!!!:
2x = 64
-------- X=32 So if your original equation was x + x+2 = 66, you just
2 2
plug in the numbers in the equation. So:
x=32, 32+ 32+2 = 66 -----> 32+34=66 ------> Your 2 consecutive numbers are 32 and 34. Hope I helped. I know it's a looong answer, but I like things to be explained :-)
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∙ 13y ago
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