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Simultaneous equations for AQA A-level Maths

Simultaneous equations

This page covers the following topics:

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.

Quadratic simultaneous equations by elimination

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.

Quadratic simultaneous equations by substitution

1

The diagram represents the graphs of xยฒ + yยฒ = 52 and y = โˆ’2x โˆ’ 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, xยฒ + (โˆ’2x โˆ’ 8)ยฒ = 52.
By expansion, xยฒ + 4xยฒ + 32x + 64 = 52.
5xยฒ + 32x + 12 = 0
Using the quadratic equation, x = (โˆ’32 ยฑ โˆš(32ยฒ โˆ’ 4 ร— 5 ร— 12)) / (2 ร— 5).
x = (โˆ’32 ยฑ 28)/10
x = โˆ’3.2 ยฑ 2.8
When x = โˆ’6, y = โˆ’2 ร— (โˆ’6) โˆ’ 8 = 4.
When x = โˆ’0.4, y = โˆ’2 ร— (โˆ’0.4) โˆ’ 8 = โˆ’7.2.
The coordinates of the points of intersection are (โˆ’6, 4) and (โˆ’0.4, โˆ’7.2).

(โˆ’6, 4) and (โˆ’0.4, โˆ’7.2)

The diagram represents the graphs of xยฒ + yยฒ = 52 and y = โˆ’2x โˆ’ 8. Use the method of substitution to find the coordinates of the points of intersection.

2

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

11xยฒ + 9yยฒ = 243
15xยฒ โˆ’ 6yยฒ = 39

Multiplying the first equation by 2 gives 22xยฒ + 18yยฒ = 486.
Multiplying the second equation by 3 gives 45xยฒ โˆ’ 18yยฒ = 117.
Adding the two equations gives 67xยฒ = 603.
Dividing both sides by 67 gives xยฒ = 9.
x = ยฑ3
Substituting this in the first equation gives 11(ยฑ3)ยฒ + 9yยฒ = 243.
9yยฒ = 144
yยฒ = 16
y = ยฑ4

x = ยฑ3, y = ยฑ4

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

11xยฒ + 9yยฒ = 243 
15xยฒ โˆ’ 6yยฒ = 39

3

The diagram shows the graphs of y = xยฒ โˆ’ 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 xยฒ โˆ’ 7x = x + 9.
Rearranging gives xยฒ โˆ’ 8x โˆ’ 9 = 0 and by factorising, (x + 1)(x โˆ’ 9) = 0.
Equating each factor to 0, x + 1 = 0 or x โˆ’ 9 = 0.
x = โˆ’1 or x = 9
When x = โˆ’1, y = โˆ’1 + 9 = 8.
When x = 9, y = 9 + 9 = 18.
The coordinates of the points of intersection are (โˆ’1, 8), (9, 18).

(โˆ’1, 8), (9, 18)

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

4

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

3y + x = 19
2xยฒ + 3yยฒ = 146

Rearranging the first equation gives x = 19 โˆ’ 3y.
Substituting this into the second equation gives 2(19 โˆ’ 3y)ยฒ + 3yยฒ = 146.
By expansion, 2(361 โˆ’ 114y + 9yยฒ) + 3yยฒ = 146.
722 โˆ’ 228y + 18yยฒ + 3yยฒ = 146
By simplifying, 21yยฒ โˆ’ 228y + 576 = 0.
Dividing every term by 3 gives 7yยฒ โˆ’ 76y + 192 = 0.
Using the quadratic equation, y = (โˆ’(โˆ’76) ยฑ โˆš((โˆ’76)ยฒ โˆ’ 4 ร— 7 ร— 192)) / (2 ร— 7) = (76 ยฑ 20)/14.
y = (76 + 20)/14 = 48/7 or y = (76 โˆ’ 20)/14 = 4
When y = 48/7, x = 19 โˆ’ 3 ร— 48 รท 7 = โˆ’11/7.
When y = 4, x = 19 โˆ’ 3 รท 4 = 7.
Then the pairs of solutions are x = โˆ’11/7, y = 48/7 and x = 7, y = 4.

x = โˆ’11/7, y = 48/7 and x = 7, y = 4

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

3y + x = 19 
2xยฒ + 3yยฒ = 146

5

Use the elimination method to solve the following simultaneous equations.

11xยฒ โˆ’ 7y = 191
13xยฒ + 14y = 493

Multiplying the first equation by 2 gives 22xยฒ โˆ’ 14y = 382.
Adding this to the second equation gives 35xยฒ = 875.
Dividing both sides by 35 gives xยฒ = 25.
x = ยฑ5
Substituting this into second equation gives 13(ยฑ5)ยฒ + 14y = 493.
14y = 168
y = 12

x = ยฑ5, y = 12

Use the elimination method to solve the following simultaneous equations. 

11xยฒ โˆ’ 7y = 191 
13xยฒ + 14y = 493

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