<p>Problem #1: </p>
<p>If a+b = 27, c= 13-b, and c= 20-a, what is a+b+c? </p>
<p>Problem #2: </p>
<p>There are 400 people at a baseball game. If there are 40 more adults than children,
and 40 more adult men than adult women, how many adults men are there at the game?</p>
<p>Is there a way to solve #2 algebraically? </p>
<p>Thanks :-)</p>
<h1>1 is 30.</h1>
<p>Set c equal to each other. You get a-b = 7.</p>
<p>Then solve for a by:</p>
<p>a+b=27
(+)
a-b=7</p>
<p>2a=34
a=17</p>
<p>17+b=27
b=10</p>
<p>plug in a or b into c to get 3.</p>
<p>17+10+3=30</p>
<p>If you are looking for an SAT-lazy way:</p>
<p>Keep in mind that they asked for a + b + c and not for the individual values. So you may suspect that there is a way to find the sum without ever solving for a, b and c.</p>
<p>If a+b = 27, c= 13-b, and c= 20-a…</p>
<p>b + c = 13</p>
<p>a + c = 20.</p>
<p>Add 'em all up: a + b + b + c + c + a = 27 + 13 + 20</p>
<p>2a + 2b + 2c = 60</p>
<p>a + b + c = 30.</p>
<p>for the first one, b=27 -a </p>
<p>then c+b = 13 which = c+27-a = 13 which = c-a = -14</p>
<p>use system of equations
c - a = -14
c + a = 20</p>
<p>subtract to cancel out c’s and you get -2a = -34, a = 17</p>
<p>with that b = 10 and c = 3. </p>
<p>a+b+c = 30</p>
<p>for the second one, c for children and a for adults.</p>
<p>c + a =400</p>
<p>c + 40 = a</p>
<p>c = a-40</p>
<p>a -40+a = 400</p>
<p>2a = 440</p>
<p>a = 220
c = 180</p>
<p>let w represent women and w +40 = men</p>
<p>w+w+40 = 220 (number of adults)</p>
<p>2w = 180</p>
<p>w = 90</p>
<p>220 - w = men</p>
<p>220 - 90 = men</p>
<p>130 = men
90 = women
180 = children</p>
<p>for total of 400.</p>