大象传媒

Solving simultaneous equations - OCRSimultaneous examples with no common coefficients

Simultaneous equations require algebraic skills to find the values of letters within two or more equations. They are called simultaneous equations because the equations are solved at the same time.

Part of MathsAlgebra

Solving simultaneous examples with no common coefficients

Some pairs of simultaneous equations may not have any common .

For example, the simultaneous equations \(3a + 2b = 17\) and \(4a - b = 30\) have no common coefficient as the coefficients of \(a\) are 3 and 4, and the coefficients of \(b\) are 2 and -1.

Remember that a common coefficient is needed, regardless of sign. This means that -1 and 1 would be seen as a common coefficient.

In examples like this, one or both equations must be multiplied to create a common coefficient.

\(\begin{array}{rrrrr} 3a & + & 2b & = & 17 \\ 4a & - & b & = & 30 \end{array}\)

Create a common coefficient for either \(a\) or \(b\). In this case, making a common coefficient for \(b\) will be easier, as \(-b\) can be doubled to create \(-2b\), which will be a common coefficient throughout the equations.

Multiply the bottom equation to create a common coefficient of \(2b\).

\(\begin{array}{rrrrr} 3a & \mathbf{+} & \mathbf{2}b & = & 17 \\ 8a & \mathbf{-} & \mathbf{2}b & = & 60 \end{array}\)

These equations can now be used to find the values of \(a\) and \(b\).

Add or subtract the two sets of equations together to solve them.

The signs in front of the common coefficients are different, so the equations should be added together:

\(\begin{array}{ccccc} 3a & + & 2b & = & 17 \\ + && + && + \\ 8a & - & 2b & = & 60 \\ = && = && = \\ 11a &&& = & 77 \\ \div 11 &&&& \div 11 \\ a &&& = & 7 \end{array}\)

Substitute the value of \(a\) into one of the original equations to find the value of \(b\).

Using \(3a + 2b = 17\) with \(a = 7\), gives \(21 + 2b = 17\), so \(b = -2\).

Therefore, the solution is \(a = 7\) and \(b = -2\).

It is a good idea to check this by using the other equation, \(4a 鈥 b = 30\): \(4a 鈥 b\) with \(a = 7\) and \(b = -2\) gives \(28 鈥 (-2) = 30\), which is correct.