 Can you get ten out of ten in this maths quiz?

# Level 7-8 Algebra - Simultaneous Equations

With all the work you've done on algebra in KS3 Maths, you will by now be familiar with equations. One type you will have come across are the simultaneous equations.

Simultaneous equations have the same values. You should be able to recognise a problem that is based on simultaneous equations – it will have two equations, and include two different variables. When a variable is the subject of one of the equations (it will start with y =… or x =...) substitute that expression into the second equation. This will produce one equation with one unknown, which can be solved. Substitute this value back into the first equation to find the second variable.

Simultaneous equations sound a bit daunting! The subject is not as fearsome as the title suggests. Play the following quiz a few times and read the helpful comments - you will soon find simultaneous equations are a doddle!

1.
Which of the following equations means exactly the same as 2x + 3y = 13?
4x + 6y = 26
6x + 4y = 26
6x + 6y = 26
6x + 6y = 13
The correct answer is double the original expression. Each term in the expression has been multiplied by two and therefore 2x becomes 4x, 3y becomes 6y and 13 becomes 26
2.
Which of the following equations means exactly the same as 7x - 5y = 17?
12x - 12y = 51
21x - 15y = 51
21x + 15y = 51
2x - 5y = 51
Here we took the original expression and multiplied each term by three. As long as we multiply each term by the SAME number, the letters in the resulting expression will always have the same value as those in the original expression
3.
Equation 1 is 2x + 3y = 19. Equation 2 is 3x + 6y = 30. How would we 'balance' one of the terms?
Add the terms in Equation 1 to the terms in Equation 2
Deduct the terms in Equation 1 from Equation 2
Multiply the terms in Equation 1 by 2
Multiply the terms in Equation 2 by 2
We would then have 4x + 6y = 38. The y's in each equation would then match and we would then say that the y's are 'balanced'
4.
Equation 1 is 8x - 3y = 17. Equation 2 is 4x + y = 21. How would we 'balance' one of the terms?
Add the terms in Equation 1 to the terms in Equation 2
Deduct the terms in Equation 1 from Equation 2
Multiply the terms in Equation 1 by 2
Multiply the terms in Equation 2 by 2
We would then have 8x + 2y = 42. The x's in each equation would be balanced
5.
We have balanced the y terms in two equations as follows: Equation 1 is 3x + 6y = 30 and Equation 2 is 4x + 6y = 38. What do we now do with them?
Add the terms in Equation 1 to the terms in Equation 2
Take the terms in Equation 1 from the terms in Equation 2
Divide Equation 1 by Equation 2
Multiply Equation 1 by Equation 2
Because the balanced terms (6y) are both positive, we take one from the other to cancel them out and we are then left with a value for x
6.
We have balanced the x terms in two equations as follows: Equation 1 is 8x - 3y = 17 and Equation 2 is 8x + 2y = 42. What do we now do with them?
Take the terms in Equation 1 from the terms in Equation 2
Add the terms in Equation 1 to the terms in Equation 2
Divide Equation 1 by Equation 2
Multiply Equation 1 by Equation 2
This time we have balanced the x terms and when we take one equation from the other we are left with a value for y. If one value for x had been positive and the other value for x had been negative we would ADD the equations together
7.
By adding (or subtracting) one equation from another we have concluded that x has a value of 8 in the equation 3x + 6y = 30. What is the value for y?
1
2
3
4
3x = 24
30 - 24 = 6
6 ÷ 6 = 1 therefore y = 1
8.
By adding (or subtracting) one equation from another we have concluded that y has a value of 5 in the equation 8x - 3y = 17. What is the value for x?
2
4
6
8
3 x 5 = 15
17 + 15 = 32
32 ÷ 8 = 4 therefore x = 4
9.
What are the values of x and y that can be derived from the following simultaneous equations: 8x - y = 13 and 12x - 3y = 15?
x = 2 and y = 2
x = 2 and y = 3
x = 3 and y = 2
x = 3 and y = 3