Simultaneous Equations Two equations, each with two unknown pronumerals, can be solved together to find one solution that satisfies both equations. There are different ways of solving simultaneous equations. The elimination method adds or subtracts the equations. The substitution method substitutes one equation into the other.

Before we discover more of these methods, there is also another way to solve simultaneous equations graphically. For example, the two equations above are y=3x+5 and y=-x+1. The simultaneous solution can be obtained by drawing the lines on a Cartesian plane, and locate the point of intersection of these two lines. Given that the two lines meet at (-1, 2) as seen, the solutions are x=-1 and y=2. Similarly, both elimination and substitution methods will yield exactly the correct results. Essentially, simultaneously solving the two equations means to obtain both pronumerals. Sometimes, there are three or more pronumerals that needs to be solved.

Elimination Method

Elimination method involves several steps:

1. Write the two equations on top of each other.
2. Decide which pronumeral to eliminate.
3. Multiply one (or both) equations so that one pronumeral could be eliminated.
4. Solve for the remaining pronumeral.
5. Substitute the solution into one of the equation to obtain the other pronumeral (which was eliminated earlier).

Best to practice!

Example:

Solve simultaneously the equations $3a+2b=5$  and $2a-b=-6$. $\begin{matrix} 3a & + & 2b & = & 5 \\ 2a & - & b & = & -6 \end{matrix}\qquad \begin{matrix} (1) \\ (2) \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \begin{matrix} 3a & + & 2b & = & 5 \\ 4a & - & 2b & = & -12 \end{matrix}\qquad \begin{matrix} (1) \\ (2)\times 2 \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \qquad \begin{matrix} \quad \\ (add\quad 2b\quad and\quad -2b\quad to\quad eliminate\quad b) \end{matrix}\\ \\ \begin{matrix} 7a & + & 0 & = & -7 \end{matrix}\\ \qquad \qquad a=-1\\ \\ Subsitute\quad a=-1\quad into\quad (1)\\ \\ 3(-1)+2b=5\\ -3+2b=5\\ 2b=8\\ b=\cfrac { 8 }{ 2 } \\ \\ b=4$

Therefore, we obtain the simultaneous solutions that a=-1 and b=4.

This method is useful for linear equations, or equations that have pronumerals with equal powers. However, with non-linear equations, substitution is the way to go.

Challenge:

A group of 39 people went to see a play. There were both adults and children in the group. The total cost of the tickets was $939, with children paying$17 each and adults paying \$29 each. How many in the group were adults and how many were children?

Substitution Method

Substitution method involves, obviously, substitution of a pronumeral into an equation.

Example:

Solve simultaneously ${ x }^{ 2 }+{ y }^{ 2 }=16$ and $3x-4y-20=0$. $\begin{matrix} { x }^{ 2 } & + & { y }^{ 2 } & = & 16 & \quad \\ 3x & - & 4y & -20 & = & 0 \end{matrix}\qquad \qquad \begin{matrix} (1) \\ (2) \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \\ From\quad (2):\quad \cfrac { 3x-20 }{ 4 } =y\qquad (3)\\ \\ Substitute\quad (3)\quad into\quad (1)\\ \\ { x }^{ 2 }+{ \left( \cfrac { 3x-20 }{ y } \right) }^{ 2 }=16\\ \\ { x }^{ 2 }+\left( \cfrac { 9{ x }^{ 2 }-120x+400 }{ 16 } \right) =16\qquad (\times \quad 16)\\ \\ 16{ x }^{ 2 }+9{ x }^{ 2 }-120x+400=256\\ 25{ x }^{ 2 }-120x+144=0\\ { (5x-12) }^{ 2 }=0\\ 5x-12=0\\ 5x=12\\ x=2.4\\ \\ Substitute\quad x=2.4\quad into\quad (3)\\ \\ y=\cfrac { 3(2.4)-20 }{ 4 } \\ \\ y=-3.2$

Therefore, the solutions are x=2.4 and y=-3.2.

Equation With 3 Unknown Variables

Solve simultaneously can be stop a-b+c=7a+2b-c=-4 and 3a-b-c=3. $\begin{matrix} a & - & b & + & c & = & 7 \end{matrix}\qquad \quad (1)\\ \begin{matrix} a & + & 2b & - & c & = & -4 \end{matrix}\quad \quad (2)\\ \begin{matrix} 3a & - & b & - & c & = & 3 \end{matrix}\quad \quad \quad (3)\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$ $\begin{matrix} a & - & b & + & c & = & 7 \end{matrix}\\ \begin{matrix} a & + & 2b & - & c & = & -4 \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ _{ _{ _{ _{ _{ _{ \quad } } } } } }(add\quad to\quad elminate\quad c)\\ \begin{matrix} 2a & + & b & \quad & \quad & = & 3 \end{matrix}\qquad (4)$ $\begin{matrix} a & - & b & + & c & = & 7 \end{matrix}\\ \begin{matrix} 3a & - & b & - & c & = & 3 \end{matrix}\\ _{ _{ _{ _{ _{ _{ _{ _{ \quad } } } } } } } }\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \quad (add\quad to\quad eliminate\quad c)\\ \begin{matrix} 4a & - & 2b & \quad & \quad & = & 10 \end{matrix}\quad \quad (5)$ $Add\quad (4)+(5)\\ \\ \begin{matrix} 2a & + & b & = & 3 \end{matrix}\\ \begin{matrix} 4a & - & 2b & = & 10 \end{matrix}\qquad (divide\quad by\quad 2)\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$ $\begin{matrix} 2a & + & b & = & 3 \end{matrix}\\ \begin{matrix} 2a & - & b & = & 5 \end{matrix}\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \quad (add\quad to\quad eliminate\quad b)\\ \\ 4a=8\\ a=2$ $Substitute\quad a=2\quad into\quad (4)\\ \\ 2(2)+b=3\\ 4+b=3\\ b=-1\\ \\ Substitute\quad a=2\quad and\quad b=-1\quad into\quad (1)\\ \\ 2-(-1)+c=7\\ 2+1+c=7\\ 3+c=7\\ c=4\\ \\ Therefore\quad the\quad solutions\quad are:\quad a=2,\quad b=-1,\quad c=4$

Watch this video to learn how to solve simultaneous equations: