### Video Transcript

Solve the simultaneous equations
for π₯ and π¦: seven π₯ plus two π¦ equals negative three and π₯ minus three π¦
equals 16.

We can solve a pair of linear
simultaneous equations by elimination or by substitution. We are firstly gonna look at the
method of elimination. Our first step is to make the
coefficients of π₯ or the coefficients of π¦ the same. This means that we want the same
number in front of both π₯ terms or both π¦ terms. We could multiply equation two by
seven, as both coefficients of π₯ would then be seven.

However, in this case, weβre going
to multiply the top equation by three and the bottom equation by two. Multiplying the top equation by
three gives us 21π₯ plus six π¦ equals negative nine. And multiplying the second equation
by two gives us two π₯ minus six π¦ equals 32.

As the signs in front of the six
π¦s are different, weβre going to add the two equations. Six π¦ plus minus six π¦ is equal
to zero. Adding the π₯ terms gives us 23π₯,
as 21π₯ plus two π₯ is 23π₯. Negative nine plus 32 is equal to
23. Dividing both sides of this
equation by 23 gives us a value of π₯ equal to one.

We now need to substitute π₯ equals
one back into one of the original equations, either equation one or equation
two. In this case, weβre going to
substitute π₯ equals one into equation one. This gives us seven multiplied by
one plus two π¦ equals negative three.

Simplifying this leaves us with
seven plus two π¦ equals negative three. Subtracting seven from both sides
of this equation gives us two π¦ equals negative 10. And finally, dividing both sides by
two gives us a value of π¦ equal to negative five. This means that the solution to the
simultaneous equations by elimination is π₯ equals one and π¦ equals negative
five.

Weβre now going to look at solving
the simultaneous equations by substitution. Our first step here is to make
either π₯ or π¦ the subject of one of the equations. In this case, weβre going to make
π₯ the subject of equation two.

Adding three π¦ to both sides of
equation two gives us π₯ is equal to 16 plus three π¦. We now need to substitute this into
equation one. This gives us seven multiplied by
16 plus three π¦ plus two π¦ equals negative three.

Expanding or multiplying out the
brackets gives us 112 plus 21π¦, as seven multiplied by 16 is 112 and seven
multiplied by three π¦ is 21π¦. Grouping our like terms gives us
112 plus 23π¦ equals negative three. Subtracting 112 from both sides of
this equation gives us 23π¦ equals negative 115. And finally, dividing both sides of
this equation by 23 gives us an answer of π¦ equals negative five.

We now need to substitute π¦ equals
negative five back into equation three to calculate π₯. π₯ is equal to 16 plus three
multiplied by negative five. Three multiplied by negative five
is negative 15. So weβre left with π₯ equals 16
minus 15. Therefore, π₯ is equal to one.

Solving the simultaneous equations
by substitution gave us the same answers: π₯ equals one and π¦ equals negative
five. We can check these answers by
substituting the values into equation one, equation two, or equation three.