x y = 2 x + 2y = 14 x = 2, y = 0 x = 3, y = 1 x = 4, y = 2 x = 5, y = 3 x = 6, y = 4 x = 7, y = 5 x = 0, y = 7 x = 2, y = 6 x = 4, y = 5
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1 List six positive integer solutions for each of these equations and comment on your results. Two have been done for you. x y = x + y = 4 x =, y = 0 x = 3, y = x = 4, y = x = 5, y = 3 x = 6, y = 4 x = 7, y = 5 x = 0, y = 7 x =, y = 6 x = 4, y = 5 x = 6, y = 4 x = 8, y = 3 x = 0, y = If we considered all solutions, not just positive integers, there would be an infinite number of answers. However, there is one solution which is correct for both of these equations simultaneously. teachitmaths.co.uk
2 We can see this clearly on a graph: x y = x + y = 4 x =, y = 0 x = 3, y = x = 4, y = x = 5, y = 3 x = 6, y = 4 x = 7, y = 5 x = 0, y = 7 x =, y = 6 x = 4, y = 5 x = 6, y = 4 x = 8, y = 3 x = 0, y = teachitmaths.co.uk
3 Simultaneous linear equations Solving two simultaneous equations involves finding the unique solution which satisfies both of them. There are two algebraic methods you could use: Elimination method go Substitution method go teachitmaths.co.uk
4 Elimination method Comparing the coefficients between each equation tells us how difficult the simultaneous equations will be to solve. 6x + y = 0 3x + y = Beginner Matching coefficients go 3p + 4q = 4 p + 5q = 9 3x + 8y = 3 x + 7y = 3 Intermediate One coefficient is a multiple of the other Expert Coefficients are not multiples go go teachitmaths.co.uk
5 Elimination method beginner 6x + y = 0 3x + y = 3x = 9 x = y = y =. Label the equations.. Compare the coefficients both y terms have a coefficient of. 3. Compare the signs both y terms are positive; subtract from to eliminate y. 4. Solve for x. 5. With an original equation, substitute x and solve for y. Why should we subtract, not add? teachitmaths.co.uk
6 Elimination method beginner 6x + y = 0 3x + y = 3x = 9 x = 3 You can check your solution by substituting the values into both solutions: 6x + y = = y = y = 3x + y = = teachitmaths.co.uk
7 Elimination method beginner 3p + q = 9 3p + 6q = 7 4q = 8 q = 3p + = 9 3p = 5 p = 5. Label the equations.. Compare the coefficients both p terms have a coefficient of Compare the signs both p terms are positive; subtract from to eliminate p. Why is this easier 4. Solve for q. than subtracting from? 5. With an original equation, substitute q and solve for p. teachitmaths.co.uk
8 Elimination method beginner x + y = x y = 6 3x = 7 x = y = y = 3. Label the equations.. Compare the coefficients both y terms have a coefficient of. 3. Compare the signs each y sign is different; add and to eliminate y. 4. Solve for x. 5. Substitute x into an original equation and solve for y. Why should we add? teachitmaths.co.uk
9 Elimination method beginner Solve these pairs of simultaneous equations: x + y = 30 x y = 4 4x + y = 3 3x + y = 8 x =, y = 4 x = 5, y = 3 x + y = 3x + y = 4 8k s = k s = x = 3, y = 5 k = 3, s = Click to show answers teachitmaths.co.uk
10 Elimination method intermediate 3p + 4q = 4 p + 5q = 9. Label the equations.. Compare the coefficients no coefficients match. 3p + 4q = 4 3p + 5q = 57 q = 33 q = 3 3. Multiply all terms in by 3, to match the p coefficients. Why is it easier to eliminate p instead of q? 4. Compare the signs signs match; subtract. p = 9 p = 4 5. Solve for q. 6. Substitute q and solve for p. teachitmaths.co.uk
11 Elimination method intermediate 5x + 4y = 3 3x y = 7 5x + 4y = 3 x 4y = 8 7x = 5 x = y = 7 y =. Label the equations.. Compare the coefficients no coefficients match. 3. Multiply all terms in by 4, to match the y coefficients. 4. Compare the signs signs are different; add. 5. Solve for x. 6. Substitute x and solve for y. teachitmaths.co.uk
12 Elimination method intermediate Solve these pairs of simultaneous equations: 3x + 7y = 47 x y = 3 3x + y = 39 5x + 7y = 3 4x + 9y = 3 3x + y = 4 x = 4, y = 5 x =, y = 4s + 3k = 39 s + 8k = 6 x = 0, y = 9 k =, s = 9 Click to show answers teachitmaths.co.uk
13 Elimination method expert 3x + 8y = 3 x + 7y = 3 6x + 6y = 64 6x + y = 69 5y = 5 y = x + 7 = 3 x = 8. Label the equations.. Compare the coefficients no coefficients match. 3. Multiply by and by 3, to match the x coefficients. 4. Compare the signs signs match; subtract. 5. Solve for y. We could have chosen to make the y coefficients match instead how would you do this? 6. Substitute y and solve for x. teachitmaths.co.uk
14 Elimination method expert x + 9y = 4 x + 6y =. Label the equations.. Compare the coefficients no coefficients match. x + 8y = 84 6x + 8y = 36 6x = 48 x = y = y = 3. Multiply by and by 3, to match the y coefficients. 4. Compare the signs signs match; subtract. 5. Solve for x. Why did we multiply by and 3 instead of 9 and 6? 6. Substitute x and solve for y. teachitmaths.co.uk
15 Elimination method expert Solve these pairs of simultaneous equations: 9x + 3y = 30 6x y = 6 4x + 9y = 35 3x + 7y = 7 7x + y = 3 5x + 3y = 9 x = 3, y = x =, y = 8 8s + 4k = -6 3s + 6k = 3 x =, y = 3 k =, s = -3 Click to show answers teachitmaths.co.uk
16 Substitution method If one variable can be expressed easily in terms of the other, you may prefer to use the substitution method. 6x + y = 9 4x + 3y = 5 y = 9 6x suitable for the substitution method 4x + 3y = 35 7x + 5y = 59 y = ⅓(35 4x) will be tricky with the substitution method teachitmaths.co.uk
17 Substitution method 3x + y = 8 x y = 5 y = x 5 3x + (x 5) = 8 3x + 4x 0 = 8 x = 4 y = 4 5 y = 3. Label the equations.. Rewrite to give y in terms of x. 3. Substitute y into. 4. Solve for x. Remember to use brackets and watch out for signs! 5. Substitute x into an original equation and solve for y. teachitmaths.co.uk
18 Substitution method x + 3y = 9 3x y = x = 9 3y 3(9 3y) y = 57 9y y = y = 5 x = x = 4. Label the equations.. Rewrite to give x in terms of y. 3. Substitute x into. 4. Solve for y. 5. Substitute y into an original equation and solve for x. teachitmaths.co.uk
19 Substitution method Solve these pairs of simultaneous equations using the substitution method: 3x + 5y = 9 x y = 4 3x + y = 3 5x + 7y = 7 4x + 9y = 3 3x + y = 4 x = 3, y = x =, y = s + 4k = 6 5s + 3k = 47 x = 4, y = k = -, s = 0 Click to show answers teachitmaths.co.uk
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