Simultaneous Equations

Let's go back to our original equations:

$2x+y=5$

$x-y=1$

We can see that if we add each side of the two equations together, we'll be able to easily eliminate y. (You can add or subtract the equations depending on the situation.) Let's try it!

$2x+y+x-y=3x$

$5+1=6$

This leaves us with the single equation:

$3x=6$

$\Rightarrow x=2$

Now we can substitute this into one of the original equations to find y:

$2-y=1$

$\Rightarrow y=1$

Finally, it's important to always check your work by substituting these values into the other equation. This is crucial to help you spot mistakes you may have made.

$2x+y=5$

$2\left(2\right)+1=5$

Solve $x+4y=11$ $x-y=1$

To begin with you can subtract each side of the one equation from one another in order to eliminate the x;

$(x+4y)-(x-y)=5y$

$11-1=10$

Now you can work with the answers to form a new equation to solve for y;

$5y=10$

$y=2$

This answer can now be substituted into one of the original equations to solve for x;

$x-y=1$

$x-2=1$

$x=1+2=3$

Finally, don't forget to check your answers by substituting them into one of the equations;

$x+4y=11$

$3+4\left(2\right)=11$

Solve $4x+2y=44$ $7x+4y=83$

To do this you first need to make two of the coefficients the same, to do this we can multiply the first equation by 2, which gives us these equations;

$8x+4y=88$

$7x+4y=83$

Now you can subtract one equation from the other to eliminate y and solve for x;

$8x+4y-7x-4y=x$

$88-83=5$

$x=5$

Now you can input your x into one of the equations to solve for y;

$4x+2y=44$

$4\left(5\right)+2y=44$

$20+2y=44$

$y=\frac{24}{2}$

$y=12$

Don't forget it is possible to check your answers by substituting both answers into the equation and checking that it works;

$4\left(5\right)+2\left(12\right)=44$

1) $5x+y=28$

2) $y=2x$

To use the substitution method you can begin by substituting equation 2 into the y of equation 1 to help you find x;

$5x+y=28$

$5x+2x=28$

$x=\frac{28}{7}$

$x=4$

You can now substitute 4 into one of the equations to find y;

$y=2x$

$y=2\left(4\right)$

$y=8$

1) $2x+y=5$

2) $x-y=1$

In this case we need to rearrange equation 2 in order to get it in terms of y, doing this will allow us to substitute it into equation 1 and solve for x;

3) $y=x-1$

4) $2x+(x-1)=5$

$\Rightarrow 3x-1=5$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

The last step remains the same: we just use this value in the second equation, which gives us y = 1 as before. Then make sure to check your answer!

1) $y-3x=6$

2)$y=x^2+2x$

On a graph, these equations would look like this:

The first thing we need to do in this case is rearrange the first equation. The key is to make them both equal y, (ie equal to the same thing), because that means we can make them both equal to each other.

3) $y=3x+6$

$\Rightarrow x^2+2x=3x+6$

$\Rightarrow x^2-x-6=0$

Now we have a quadratic equation that you can solve using your preferred method. In this example we're going to use factorization . (If you need some practice with this, check out our article on Factoring Quadratic Expressions!)

$x^2-x-6=0$

$\Rightarrow (x-3)(x+2)$

$\Rightarrow x=3orx=-2$

As you can see, we have two possible values for x. This means we need to substitute them each back in to find two pairs of solutions for our simultaneous equations:

1) $y=3\left(3\right)+6=15$ or 1)$y=3(-2)+6=0$

This gives us the pairs

$x=3,y=15$or $x=-2,y=0$

check:

2) $y=x^2+2x$

$y={\left(3\right)}^2+2\left(3\right)=9+6=15$

$y={(-2)}^2+2(-2)=4-4=0$

Ali buys 2 toffees and 3 gumballs. Bea buys 3 toffees and 2 gumballs. Ali's total comes to £0.10 and Bea pays £0.15. How much are toffees and gumballs?

First, we need to identify the variables. In this case they are toffees and gumballs. We can see that 2 toffees and 3 gumballs cost 10p, and 3 toffees and 2 gumballs cost 15p. Using this information we can write the following equations, where g represents the gumballs and t represents the toffees:

1) $3g+2t=10$

2) $2g+3t=15$

You can see these are actually one of the sets of simultaneous equations from before!