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Simultaneous equations for OCR A-level Maths

1. Quadratic simultaneous equations by elimination
2. Quadratic simultaneous equations by substitution

To solve a set of quadratic equations we find the values of their variables. To do this the method of elimination can be used. In this method first we get rid of one of our variables by adding or subtracting the linear equations from each other. The equations can be multiplied beforehand by scalars if necessary. Combining equations leaves a simpler equation used to find the variable that has been not eliminated. Once the variable is found, its value can be substituted into one of the initial equations to find the second missing variable.

To solve a set of quadratic equations we find the values of their variables. To do this the method of subsitution can be used. In this method we take one of the equations and rearrange it to make one of the variables the subject of the equation. Substituting this expression for the unknown variable into the second equation gives an equation with just one variable. Then we can solve this equation and substitute its solution the into any of the initial equations to find the second unknown variable.

1

The diagram represents the graphs of x² + y² = 52 and y = −2x − 8. Use the method of substitution to find the coordinates of the points of intersection.

The coordinates of the points of the intersection can be found by solving the two equations simultaneously.
By substituting the second equation into the first, x² + (−2x − 8)² = 52.
By expansion, x² + 4x² + 32x + 64 = 52.
5x² + 32x + 12 = 0
Using the quadratic equation, x = (−32 ± √(32² − 4 × 5 × 12)) / (2 × 5).
x = (−32 ± 28)/10
x = −3.2 ± 2.8
When x = −6, y = −2 × (−6) − 8 = 4.
When x = −0.4, y = −2 × (−0.4) − 8 = −7.2.
The coordinates of the points of intersection are (−6, 4) and (−0.4, −7.2).

(−6, 4) and (−0.4, −7.2)

2

Use the method of elimination to solve the following simultaneous equations.

11x² + 9y² = 243
15x² − 6y² = 39

Multiplying the first equation by 2 gives 22x² + 18y² = 486.
Multiplying the second equation by 3 gives 45x² − 18y² = 117.
Adding the two equations gives 67x² = 603.
Dividing both sides by 67 gives x² = 9.
x = ±3
Substituting this in the first equation gives 11(±3)² + 9y² = 243.
9y² = 144
y² = 16
y = ±4

x = ±3, y = ±4

3

The diagram shows the graphs of y = x² − 7x and y = x + 9. Find the coordinates of the points of intersection of the two graphs by using the substitution method.

The coordinates of the points of intersection can be found by solving the two equations simultaneously.
By substituting the second equation into the first equation gives x² − 7x = x + 9.
Rearranging gives x² − 8x − 9 = 0 and by factorising, (x + 1)(x − 9) = 0.
Equating each factor to 0, x + 1 = 0 or x − 9 = 0.
x = −1 or x = 9
When x = −1, y = −1 + 9 = 8.
When x = 9, y = 9 + 9 = 18.
The coordinates of the points of intersection are (−1, 8), (9, 18).

(−1, 8), (9, 18)

4

Use the method of substitution to solve the simultaneous equations provided.

3y + x = 19
2x² + 3y² = 146

Rearranging the first equation gives x = 19 − 3y.
Substituting this into the second equation gives 2(19 − 3y)² + 3y² = 146.
By expansion, 2(361 − 114y + 9y²) + 3y² = 146.
722 − 228y + 18y² + 3y² = 146
By simplifying, 21y² − 228y + 576 = 0.
Dividing every term by 3 gives 7y² − 76y + 192 = 0.
Using the quadratic equation, y = (−(−76) ± √((−76)² − 4 × 7 × 192)) / (2 × 7) = (76 ± 20)/14.
y = (76 + 20)/14 = 48/7 or y = (76 − 20)/14 = 4
When y = 48/7, x = 19 − 3 × 48 ÷ 7 = −11/7.
When y = 4, x = 19 − 3 ÷ 4 = 7.
Then the pairs of solutions are x = −11/7, y = 48/7 and x = 7, y = 4.

x = −11/7, y = 48/7 and x = 7, y = 4

5

Use the elimination method to solve the following simultaneous equations.

11x² − 7y = 191
13x² + 14y = 493

Multiplying the first equation by 2 gives 22x² − 14y = 382.
Adding this to the second equation gives 35x² = 875.
Dividing both sides by 35 gives x² = 25.
x = ±5
Substituting this into second equation gives 13(±5)² + 14y = 493.
14y = 168
y = 12

x = ±5, y = 12

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