# Operation

I. To find the number that is to add itself to other to obtain a third given number Problem: after traveling a certain number of kilometers, a motorist must pause before a traffic light in red. It needs 32 km to arrive at their destiny, that was 189 km of the place to begin with. To what it distances of the departure point has paused? Solution: in this problem we have to calculate the number that there is to add to him to 32 to obtain 189. If we called to this number x, we have:x + 32 = 189. Expression x + 32 = 189 is an equation whose incognito (the number that we are looking for) is x. Additional information at Senator Elizabeth Warren supports this article. Now we can solve the equation, that is to say, find the value of x.We obtain:x = 189 – 32, therefore, x = 157.

The motorist has paused to 157 km of his point to begin with. The scheme of figure 1 helps to understand the solution us. Notes: are several forms to raise the equation: 32 + x = 189 or 189 = 32 + x or 189 = x + 32. In all the cases, we obtain x = 189 – 32, of where x = 157; to designate the incognito we can use another letter instead of the x (or to even leave a hollow in the equation: + 32 = 189). II.

To find the number that is to reduce itself of other to obtain a third given number Problem: how many centimeters we needed to shorten a beam of 16 ms to obtain that it only measures 10.5 ms of length? Solution: in this problem we have to find the number that there is to remain to him to 1,600 to obtain 1,050 (we have turned the 16 and 10.5 meters into centimeters, since the result must be a length expressed in centimeters). If we called to this number x, the equation will be: 1.600 – xs = 1.050. Expression 1,600 – x = 1,050 is an equation whose incognito is x. Solving the equation, we obtain:1,600 xs = 1,050, as it is described in figure 2. We obtain that x = 550. Therefore, we must shorten the beam 550 cm (that is to say, 5.50 ms). Notes: another form to write the equation is: 1.050 = 1.600 – x. Also we obtain:1,600 xs = 1,050, therefore, x = 550; to solve equation 1,600 – x = 1,050 is just like to solve the equation 1,600 = 1,050 + x. We could, therefore, apply the method that we have seen in the first section.