LO I can solve simultaneous equations Date 08032021

LO: I can solve simultaneous equations. Date 08/03/2021 Word Bank: 1. 2. 3. 4. 5. 6. Increase 2800 kg by 24% Work out the value of: 5 x - 2 y When x = 3 and y = -4 Factorise: 2 b 2 – 2 b Work out the LCM of 6 and 9? Round off 0. 67 to one s. f The equation of a line is y=x + 7 Give the gradient and y-intercept Challenge task: 5. 7 is rounded to one decimals place. Write down the maximum possible it could have been. 5. 75 Simultaneous Integer

LO: I can solve simultaneous equations. Date 08/03/2021 Word Bank: 1. 2. 3. 4. 5. 6. Increase 2800 kg by 24% 3472 kg Work out the value of: 5 x - 2 y When x = 3 and y = -4 23 Factorise: 2 b 2 – 2 b 2 b(b-1) Work out the LCM of 6 and 9? 18 Round off 0. 67 to one s. f 0. 7 The equation of a line is y=x + 7 Give the gradient and y-intercept 1 & 7 Challenge task: 5. 7 is rounded to one decimals place. Write down the maximum possible it could have been. 5. 75 Simultaneous Integer

Simultaneous Equations LO: 08/03/2021 I can : I can solve simultaneous equations. Is to: Find solutions to equations. Keywords Developing Integer Simultaneous Is to: Solve simultaneous equation using the elimination method. Solve simultaneous equation using the substitution method. Emerging Secure

Simultaneous Equations Skills Check 08/03/2021

Simultaneous Equations 08/03/2021

Simultaneous Equations List six positive integer solutions for each of these equations. Two have been done for you. x–y=2 x + 2 y = 14 x = 2, y = 0 x = 0, y = 7 x = 3, y = 1 x = 2, y = 6 x = 4, y = 2 x = 4, y = 5 x = 5, y = 3 x = 6, y = 4 x = 8, y = 3 x = 7, y = 5 x = 10, y = 2 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. 08/03/2021

Simultaneous Equations We can see this clearly on a graph: x–y=2 x + 2 y = 14 x = 2, y = 0 x = 0, y = 7 x = 3, y = 1 x = 2, y = 6 x = 4, y = 2 x = 4, y = 5 x = 5, y = 3 x = 6, y = 4 x = 8, y = 3 x = 7, y = 5 x = 10, y = 2 08/03/2021

Simultaneous Equations 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 08/03/2021

Simultaneous Equations 08/03/2021 Elimination method Comparing the coefficients between each equation tells us how difficult the simultaneous equations will be to solve. 6 x + y = 20 3 x + y = 11 Beginner Matching coefficients go 3 p + 4 q = 24 p + 5 q = 19 Intermediate One coefficient is a multiple of the other go 3 x + 8 y = 32 2 x + 7 y = 23 Expert Coefficients are not multiples go

Simultaneous Equations Elimination method – beginner 6 x + y = 20 3 x + y = 11 3 x 1 2 =9 x=3 3 × 3 + y = 11 y=2 1. Label the equations. 2. Compare the coefficients – both y terms have a coefficient of 1. 3. Compare the signs – both y terms are positive; subtract 2 from 1 to eliminate y. 2 4. Solve for x. Why should we subtract, not add? 5. With an original equation, substitute x and solve for y. 08/03/2021

Simultaneous Equations Elimination method – beginner 6 x + y = 20 3 x + y = 11 3 x 1 2 6 x + y = 20 6 × 3 + 2 = 20 =9 x=3 3 × 3 + y = 11 y=2 You can check your solution by substituting the values into both solutions: 2 3 x + y = 11 3 × 3 + 2 = 11 1 2 08/03/2021

Simultaneous Equations Elimination method – beginner 3 p + 2 q = 19 3 p + 6 q = 27 1 2 4 q = 8 q=2 3 p + 2 × 2 = 19 3 p = 15 p=5 2 1. Label the equations. 2. Compare the coefficients – both p terms have a coefficient of 3. 3. Compare the signs – both p terms are positive; subtract 1 from 2 to eliminate p. Why is this easier than subtracting 4. Solve for q. 2 from 1 ? 5. With an original equation, substitute q and solve for p. 08/03/2021

Simultaneous Equations Elimination method – beginner 2 x + y = 21 x–y=6 3 x 1 2 = 27 x=9 2 × 9 + y = 21 y=3 1. Label the equations. 2. Compare the coefficients – both y terms have a coefficient of 1. 3. Compare the signs – each y sign is different; add 1 and 2 to eliminate y. 1 4. Solve for x. Why should we add? 5. Substitute x into an original equation and solve for y. 08/03/2021

Simultaneous Equations Elimination method – beginner Solve these pairs of simultaneous equations: x + 2 y = 30 x – 2 y = 14 4 x + y = 23 3 x + y = 18 x = 22, y = 4 x = 5, y = 3 2 x + y = 11 3 x + y = 14 8 k – s = 22 k–s=1 x = 3, y = 5 k = 3, s = 2 Click to show answers 08/03/2021

Simultaneous Equations 08/03/2021 Elimination method – intermediate 3 p + 4 q = 24 p + 5 q = 19 3 p + 4 q = 24 3 p + 15 q = 57 1 2 11 q = 33 q=3 p + 5 × 3 = 19 p=4 1. Label the equations. 2. Compare the coefficients – no coefficients match. 3. Multiply all terms in 2 by 3, to match the p coefficients. Why is it easier to eliminate p instead of q? 4. Compare the signs – signs match; subtract. 2 5. Solve for q. 6. Substitute q and solve for p.

Simultaneous Equations 08/03/2021 Elimination method – intermediate 5 x + 4 y = 23 3 x – y = 7 5 x + 4 y = 23 12 x – 4 y = 28 17 x 1 2 = 51 x=3 3× 3–y=7 y=2 1. Label the equations. 2. Compare the coefficients – no coefficients match. 3. Multiply all terms in 2 by 4, to match the y coefficients. 4. Compare the signs – signs are different; add. 5. Solve for x. 2 6. Substitute x and solve for y.

Simultaneous Equations 08/03/2021 Elimination method – intermediate Solve these pairs of simultaneous equations: 3 x + 7 y = 47 2 x – y = 3 4 x + 9 y = 13 3 x + y = 4 x = 4, y = 5 x = 1, y = 1 3 x + y = 39 5 x + 7 y = 113 4 s + 3 k = 39 2 s + 8 k = 26 x = 10, y = 9 k = 1, s = 9 Click to show answers

Simultaneous Equations Elimination method – expert 3 x + 8 y = 32 2 x + 7 y = 23 6 x + 16 y = 64 6 x + 21 y = 69 1 2 2. Compare the coefficients – no coefficients match. 1 3. Multiply 1 by 2 and 2 by 3, to match the x coefficients. 2 5 y = 5 y=1 2 x + 7 = 23 x=8 1. Label the equations. We could have chosen to make the y coefficients match instead – how would you do this? 4. Compare the signs – signs match; subtract. 2 5. Solve for y. 6. Substitute y and solve for x. 08/03/2021

Simultaneous Equations Elimination method – expert 11 x + 9 y = 42 2 x + 6 y = 12 22 x + 18 y = 84 6 x + 18 y = 36 16 x 1 2 2. Compare the coefficients – no coefficients match. 1 3. Multiply 1 by 2 and 2 by 3, to match the y coefficients. 2 = 48 x=3 2 × 3 + 6 y = 12 y=1 1. Label the equations. Why did we multiply by 2 and 3 instead of 9 and 6? 4. Compare the signs – signs match; subtract. 2 5. Solve for x. 6. Substitute x and solve for y. 08/03/2021

Simultaneous Equations Elimination method – expert Solve these pairs of simultaneous equations: 9 x + 3 y = 30 6 x – 2 y = 16 7 x + 2 y = 23 5 x + 3 y = 29 x = 3, y = 1 x = 1, y = 8 4 x + 9 y = 35 3 x + 7 y = 27 8 s + 4 k = -16 3 s + 6 k = 3 x = 2, y = 3 k = 2, s = -3 Click to show answers 08/03/2021

Simultaneous Equations Substitution method If one variable can be expressed easily in terms of the other, you may prefer to use the substitution method. 6 x + y = 19 4 x + 3 y = 15 y = 19 – 6 x 4 x + 3 y = 35 7 x + 5 y = 59 y = ⅓(35 – 4 x) suitable for the substitution method will be tricky with the substitution method 08/03/2021

Simultaneous Equations Substitution method 3 x + 2 y = 18 2 x – y = 5 y = 2 x – 5 3 x + 2(2 x – 5) = 18 3 x + 4 x – 10 = 18 x=4 y=2× 4 – 5 y=3 1 1. Label the equations. 2 2 1 2 2. Rewrite 2 to give y in terms of x. . 3. Substitute y into 1 4. Solve for x. Remember to use brackets and watch out for signs! 5. Substitute x into an original equation and solve for y. 08/03/2021

Simultaneous Equations Substitution method x + 3 y = 19 3 x – 2 y = 2 x = 19 – 3 y 3(19 – 3 y) – 2 y = 2 57 – 9 y – 2 y = 2 y=5 x = 19 – 3 × 5 x=4 1 1. Label the equations. 2 1 2 2. Rewrite 1 to give x in terms of y. 3. Substitute x into 2 . 4. Solve for y. 1 5. Substitute y into an original equation and solve for x. 08/03/2021

Simultaneous Equations Substitution method Solve these pairs of simultaneous equations using the substitution method: 3 x + 5 y = 19 2 x – y = 4 4 x + 9 y = 13 3 x + y = 4 x = 3, y = 2 x = 1, y = 1 3 x + y = 13 5 x + 7 y = 27 s + 4 k = 6 5 s + 3 k = 47 x = 4, y = 1 k = -1, s = 10 Click to show answers 08/03/2021