I have this systems of equation and I'm suppose to solve it using substitution. 2a-b=17 3a+4b=-13 I have the first part of the answer: a=(81/5). Then, I plug it into the other equation: 2a-b=17, which becomes 2(81/5)-b=17. So, I get (162/5)-b=17. This is the part where I'm completely lost. Do I subtract (162/5) from both sides, or do I multiply both sides by 5 and then subtract 162?

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I have this systems of equation and I'm suppose to solve it using substitution. 2a-b=17 3a+4b=-13 I have the first part of the answer: a=(81/5). Then, I plug it into the other equation: 2a-b=17, which becomes 2(81/5)-b=17. So, I get (162/5)-b=17. This is the part where I'm completely lost. Do I subtract (162/5) from both sides, or do I multiply both sides by 5 and then subtract 162?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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you just need to move (162/5) from left to the right so, become -b=17-(162/5) and then -b= -(77/5) so, b will be 77/5. and done

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