Simultaneous equations – Cont’d
WE WILL complete our review of simultaneous equations this week by looking at the solutions to some of the practice examples that were given for homework.
You are reminded that the two methods of note are the elimination and substitution methods. By now, you should know them well. Solve simultaneously: (Using the elimination method) 2n + m = 3 . . .(1) 5n - 2m = 12 . . .(2) Multiply equation (1) by 2 4n + 2m = 6. (3) Add equations (2) and (3) 9n = 18 n = 18/9 = 2 Substituting n = 2 into equation (1): 2x2+m=3 4+m=3 m = 3 - = -1 m=-1 Answer: n = 2, m = - 1.
Solve simultaneously: 2x = 11 + 3y . . . (1) x + 2y + 12 = 0 . . . (2) Using the substitution method: From equation (1), x = 11 + 3y/2 Substituting into equation (2): 11 + 3y/2 + 2y +12 = 0. Multiply all terms by 2 to clear the denominator: 2 x 11 + 3y/2 + 2 x 2y + 2 x 12 = 2 x 0. 11 + 3y + 4y + 24 = 0 11 + 7y + 24 = 0 7y = - 24 -11 = - 35 y = - 35 /7 = - 5 Substituting into equation (1): x = 11 + 3y/2 = 11 + 3 x - 5/2 = 11 - 15/2 x = - 4/2 = -2 Answer: x = -2, y = -5. You may solve the following on your own: x+y=7 2x + y = 10 3x - 4y = 32 5x + 2y = 10 -x+y=1 3x + y = 9
Please continue to practise on your own, especially the word problems where reasoning is required.
EXAMPLE
Four mangoes and two pears cost $24, while two mangoes and three pears cost $16. Write a pair of simultaneous equations in x and y to represent the information given above and solve them.
SOLUTION
Let the cost of mangoes be x and the cost of pears be y. From the above : 4x + 2y = 24 (1) 2x + 3y = 16 (2) Multiply equation (2) by 2. 4x + 6y = 32 (3) Equation (3) - (1) 4y = 8, y = 2 Substituting in (2) 2x + 6 = 16, 2x = 16 - 6 2x = 10, x = 5 Answer: x = $5 and y = $2
EXAMPLE
The width of a rectangle is 7cm less than its length. If its perimeter is 50cm, calculate its dimensions.
SOLUTION
Let the dimensions of the rectangle be x and y. The equations are: x = y - 7 (1) 2x + 2y = 50 (2) Rearranging equation (1) multiply (3) by 2 x - y = - 7 (3) 2x - 2y = - 14 (4) Adding (2) + (4) 4x = 36 x = 36 / 4 = 9 Substituting in (1) 9=y-7 y = 9 + 7 = 16 The dimensions are 9cm and 16cm. Let us attempt together the following: (e) If x is a real number, solve, 8 - x 5x + 2
SOLUTION
If 8 - x 5x + 2 then 8 - 2 5x + x 6x 6x > 1 Please observe the sign.
The values of x are all real numbers equaling 1 and greater.
NB: If in manipulating the in-equation, you divide both sides by - 1, then you have to reverse the in-equality sign.
For example, x > - 7, if you divide by - 1, then, - x < 7 (f) If x is a real number, solve, 2n2 - 3n - 20 0 Solution : Consider the equation, 2n2 - 3n - 20 = 0 Solving using factorising2n2 - 3n - 20 = (n - 4 )(2n + 5 ) Since, (n - 4 )(2n + 5 ) = 0, then(n - 4 ) = 0, n = 4 Or (2n + 5 ) = 0 n = - 5/2
There are two possible answers, n - 5/2 and n 4, or - 5/2 n 4
Since 2n2 - 3n - 20 > 0, n = 0 does not satisfy the in-equation, ie - 20 > 0
If you attempt n = - 3, then 7 > 0 which is satisfied.
The correct range or solution is: n - 5/2 and n > 4, NB: n = - 3, is in the range of the answer.
SOLUTION OF QUADRATIC EQUATIONS EXTRACT FROM SYLLABUS
Specific Objectives Solve quadratic equation. Solve word problems . Content
Linear equations, linear inequalities, two simultaneous linear equations and quadratic equations. The following are the methods which are commonly used at this level. Factorisation Graphs Formula method
POINTS TO NOTE
Quadratic equations are expressed in the form ax2 + bx + c = 0, where a, b and c are constants.
The factorisation method is used if, and only if, the expression ax2 + bx + c can be factorised. The solution is represented by two values of x. Given the equation x2 + 8x + 12 = 0, then by factorising the left-hand side, you get . (x + 2 )( x + 6 ) = 0. Continuing the method: If (x + 2 )( x + 6 ) = 0 then (x + 2) = 0, that is x = - 2 OR (x + 6) = 0, that is x = - 6. Solutions are x = -2 or x = - 6. Be reminded that:
The solutions of the equation are the values which satisfy the equation.
These can be checked by substitution as follows:
If x2 + 8x + 7 = 0, then if x = -1, 1 -8 + 7 = 0. Similarly, where x = -7, then 49 -56 + 7 = 0. The equation is satisfied by both values. We shall now look at some other examples. 1. Solve 3x2 - 7x -6 = 0
Using factorisation: If 3x2 - 7x -6 = 0 (3x + 2) (x - 3) = 0 3x + 2 = 0, that is, 3x - 2 x = - 2/3 When x - 3 = 0, x=3 Answer: x = -2/3 or 3 Solve the following quadratic equations. x2 + 4x + 3 = 0 x2 - 8x + 7 = 0 x2 - 3x - 10 = 0 6x2 - x - 15 = 0 2x2 - x - 3 = 0 x2 + 4x = 12 3x2 - 5x - 2 = 0 8a2 - 1 = 2a Solve the simultaneous equations: 3x - 1/2y = 4 9x + 2y = -2 I urge you to find other examples in your textbooks and past papers and do them.