MATHS

Asked by MehakSo the question shows the ratio a+b-c : b+c-a : a+c-b and it says that its equal to the ratio 5:6:7 This means that: a+b-c = 5 b+c-a = 6 a+c-b = 7 Lets label these equations as 1, 2 and 3 1) a+b-c = 5 2) b+c-a = 6 3) a+c-a = 7 Now we are going to work with it like simultaneous equations So lets take 1) and 2) first a+b-c = 5 b+c-a = 6 _________ 2b=11 b=11/2 We worked out the value of b by adding the two simultaneous equations together and then solving for b Now if you take 1) and 3) a+b-c = 5 a+c-b = 7 _________ 2a=12 a=6 Like how we worked out value of b same thing to work out value of a ....add simultaneous equations and solve for a. Now you can either substitute the value of a and b into one of the equations to work out c or you can do another simultaneous equation with 2) and 3) Simultaneous equation: b+c-a = 6 a+c-b = 7 _________ 2c =13 c=13/2 Substitution method: Im going to take 3) this equation a+c-b = 7 we worked out that a=6 and b=11/2 So this would be 6+c- 11/2 = 7 If you solve it by adding 11/2 to 7 and taking away 6 . You will get c =13/2 So your values are as such a = 6 , b =11/2 and c = 13/2 If you dont want fractions then multiply all values by 2 to give you a= 12, b=11 and c=13

First I would separate out the equations so it's easier to break down for yourself: a+b-c = 5 b+c-a = 6 a+c-b = 7 There are a few ways to solve this, today lets use 'elimination'. This is where we will add two equations together and try to 'cancel out' certain letters or values. let's look for a two equations which 'cancel out' something if we added them together. In the first equation there is a '-c' so lets add it to an equation that has '+c' (take the second equation for example) So line up the two equations and rearrange them so that the values are neatly aligned: a+b-c = 5 -a+b+c = 6 <-this was originally "b+c-a = 6" If you look closely, it looks like both the 'a's and the 'c's will cancel out, now treat this like a simple addition, and leave the equals sign in! +a +b -c = 5 -a +b +c = 6 + --------------- 0 + 2b + 0 = 11 so now we know that 2b = 11, we know that "b = 11/2" (or 5.5) Let's go back and see if we can use this method to find another letter, we can apply it to the second and third equation: b+c-a = 6 a+c-b = 7 the 'b's and 'a's should cancel out -a +b +c = 6 +a -b +c = 7 + --------------- 0 + 0 + 2c = 13 now that we know 2c = 13, we know that c = 13/2 (or 6.5) using 'b = 5.5' and 'c = 6.5' we can substitute these values into any of the equations to find 'a'. b + c - a = 6 5.5 + 6.5 - a = 6 12 - a = 6 12 = 6 + a 6 = a So: a = 6 b = 5.5 c = 6.5

So the question shows the ratio a+b-c : b+c-a : a+c-b and it says that its equal to the ratio 5:6:7 This means that: a+b-c = 5 b+c-a = 6 a+c-b = 7 Lets label these equations as 1, 2 and 3 1) a+b-c = 5 2) b+c-a = 6 3) a+c-a = 7 Now we are going to work with it like simultaneous equations So lets take 1) and 2) first a+b-c = 5 b+c-a = 6 _________ 2b=11 b=11/2 We worked out the value of b by adding the two simultaneous equations together and then solving for b Now if you take 1) and 3) a+b-c = 5 a+c-b = 7 _________ 2a=12 a=6 Like how we worked out value of b same thing to work out value of a ....add simultaneous equations and solve for a. Now you can either substitute the value of a and b into one of the equations to work out c or you can do another simultaneous equation with 2) and 3) Simultaneous equation: b+c-a = 6 a+c-b = 7 _________ 2c =13 c=13/2 Substitution method: Im going to take 3) this equation a+c-b = 7 we worked out that a=6 and b=11/2 So this would be 6+c- 11/2 = 7 If you solve it by adding 11/2 to 7 and taking away 6 . You will get c =13/2 So your values are as such a = 6 , b =11/2 and c = 13/2 If you dont want fractions then multiply all values by 2 to give you a= 12, b=11 and c=13