form and solve simultaneous equations

Transcription

1 form and solve simultaneous equations Skills that will help you to understand, work with and solve formulae and equations include the four rules of number (including working with very large and very small numbers), use of a calculator and powers and roots. You can find two unknown values by solving simultaneous equations. 1 Forming simultaneous equations Breaking it down Carefully read all the information in the question and organise it into two sets of information Form an equation for each set of information Worked example 1 Two t-shirts and a pair of jeans cost 28. Five t-shirts and a pair of jeans cost 43. Form two equations to represent this information. One set of information is about the cost of 2 t-shirts and 1 pair of jeans. The second set of information is about the cost of 5 t-shirts and 1 pair of jeans Let us use t to represent the cost of one t-shirt in and j to represent the cost of 1 pair of jeans in We can represent the first set of information as: 2t + j = 28 We can represent the second set of information as: 5t + j = 43

2 2 Solve simultaneous equations using rearrangement or elimination Method 1 Rearrangement method Breaking it down Stage 4 Look carefully at the two equations and decide which equation will be easier to rearrange into the form of x = or t = or similar Rearrange the selected equation into this form Substitute the expression from into the other equation, to give an equation which has only one unknown letter. We can solve this equation Substitute the value from into the first equation and find a value for the remaining letter Check your answers by putting both values into second equation Worked example 2 Let s use the example from Worked example 1 (above). The two equations formed are: 2t + j = 28 (We ll label this [1]) 5t + j = 43 (We ll label this [2]) In this example, both equations are equally easy to rearrange to make j the subject of the expression. Let s rearrange [1] The rearrangement results in j = 28 2t Substitute j = 28 2t into equation [2] 5t + (28 2t) = 43 5t t = 43 3t + 28 = 43 3t = t = 15 t = 5 Therefore a t-shirt costs 5 Stage 4 Substitute t = 5 into [1] (2 x 5) + j = j = 28 j = j = 18 Therefore a pair of jeans costs 18 Check your answer by putting t = 5 and j = 18 into [2] (5 x 5) + 18 = 43 This is correct; the left- and right-hand sides tally

3 Worked example 3 A 100 ml bottle of cough mixture is enough for 4 adults and 8 children or 6 adults and 2 children. What are the dosages for adults and for children? Let s use a for an adult s dose and c for a child s dose. The two equations formed are: 4a + 8c = 100 [1] and 6a + 2c = 100 [2] It will probably be easier to rearrange [2] to find make c the subject Rearranging [2] gives c = 50 3a 100! 6a c = which simplifies to 2 Substitute c = 50 3a in [1] 4a + 8 (50 3a) = 100 (remove the brackets) 4a a = 100 (collect a terms together) a = 100 (add 20a to both sides) (subtract 100 from each side) (divide both sides by 20) 400 = a 300 = 20a 15 = a Stage 4 Substitute a = 15 in [2] (6 x 15) + 2c = 100 (multiply 6 and 15 together) c = 100 (subtract 90 from both sides) 2c = 10 (divide both sides by 2) c =5 Check in [1] (4 x 15) + (8 x 5) = 100 This is correct; the left- and right-hand sides tally You could also have solved this by finding an expression for 2c and substituting this in the expression, multiplying it by 4 for 8c and continuing from above.

4 Method 2 Elimination method Breaking it down Look at the number in front of each variable (this is called the coefficient). We need to manipulate the equations so that one coefficient for one of the variables is the same numerical value in both expressions If necessary, multiply one or both equations to make one of the coefficients the same in both equations Either add or subtract the two equations to eliminate one letter The acronym STOP will help you remember whether to add or subtract the equations. If the signs of the coefficients are the Same, then Take them away; if they are Opposite, then Plus (add) Stage 4 Stage 6 Solve the equation you are left with Put the value you have found into one of the equations to find the remaining unknown (see Stage 4 in the rearrangement method) Finally, check your answer by putting both values into the remaining equation (see in the rearrangement method)

5 Worked example 4 Using the same example as before (1 & 2 above), the two equations formed are: 2t + j = 28 [1] 5t + j = 43 [2] We already have a common coefficient. We have 1j in each equation (we don t usually bother to write the 1 ) We won t need to do this stage because we already have the same coefficient for j Both the equations have a +j so the signs are the same we need to take one equation from the other. We will do [2] take away [1] so that the number of t we have left is positive 5t + j = 43 [2] 2t + j = 28 [1] Subtracting gives (5t 2t) +(j j ) = (43 28) Stage 4 This becomes 3t = 15 We can solve this to give t = 5 Put t = 5 in [1] (see Stage 4 in the rearrangement method above) j = 18 Stage 6 Check by putting the values in [2] (see in the rearrangement method above) (5 x 5) + 18 = 28 This is true; the left- and right-hand sides tally

6 Worked example 5 Using the cough medicine example (3) as before, the two equations are: 4a + 8c = 100 [1] and 6a + 2c =100 [2] If we multiply equation [2] by 4, we will get 8c. (An alternative would work if we multiplied [1] by 3 and [2] by 2. This would make 12a in both equations) Multiply all of the coefficients in [2] by 4. This gives 24a + 8c = 400 We now have the equations: 4a + 8c = 100 [1] and 24a + 8c = 400 The signs for 8c are the Same (both +), so we need to Take the equations away (remember STOP) Stage 4 (24a 4a) + (8c 8c) = a = 300 a = 15 Put a = 15 in one of the equations (we ll use [1]) (see Stage 4 in the rearrangement method above) (4 x 15) + 8c = 100 c = 5 Stage 6 Check by putting both values in to equation [2] (see in the rearrangement method above) (6 x 15) + (2 x 5) = 100 This is true; the left- and right-hand sides tally Common errors include not multiplying all of the terms when trying to make the coefficients the same; adding the equations together when they should be subtracted and vice versa which means that the no letter is eliminated; errors when rearranging the equations.