How do you solve the system of equations #7x + 5y = 30# and #x + y = 4#?

1 Answer
Jun 17, 2018

Either:
1) Substitute one equation in to the other;
2) Solve by matrices

Explanation:

The quickest way in this example is to rearrange the simpler second equation to substitute in to the first:
#y=4-x#
so
#7x+5(4-x)=30#
#7x+20-5x=30#
#2x=10#
#x=5#

Return to the second equation:
#5+y=4#
#y=-1#

So #(x,y)=(5,-1)#.

Alternative approach - matrix solution

A more general approach that becomes better when the linear equations are more complicated (e.g. when you have more equations and variables) is to write the equations as a matrix system and then use the inverse of the matrix of coefficients to obtain the answer. In this case, we would write:

#((7,5),(1,1))((x),(y))=((30),(4))#

If we now premultiply by the inverse of the matrix #((7,5),(1,1))#, then we immediately have solutions for #x# and #y#.

There are two approaches to taking the inverse. In the first you learn the formula for it:

If #A# is a 2x2 matrix, then its inverse is #A^(-1)=1/(ad-bc)((d,-b),(-c,a))#

Substituting in, we find that the inverse in this case is #1/2((1,-5),(-1,7))=((1/2,-5/2),(-1/2,7/2))#

So #(x,y)=((1/2,-5/2),(-1/2,7/2))((30),(4))=((5),(-1))#, as before.

This is immediately a more general way of approaching a system of two linear equations in two variables, but it relies on the formula above, a formula which becomes much more complex if we have more equations and variables. The second way, the usual practical way to approach such problems is to append an identity matrix to one side and then manipulate the matrix and identity matrix together in the same way until the matrix has become the identity - then the identity has become the inverse.

In this case, we might proceed

Start with:
#((7,5|1,0),(1,1|0,1))#

Multiply row 2 by 5:
#((7,5|1,0),(5,5|0,5))#

Take row 2 away from row 1 in row 1:
#((2,0|1,-5),(5,5|0,5))#

Multiply row 1 by #5/2#:
#((5,0|5/2,-25/2),(5,5|0,5))#

Take row 1 away from row 2 in row 2:
#((5,0|5/2,-25/2),(0,5|-5/2,35/2))#

Divide the whole matrix by 5:
#((1,0|1/2,-5/2),(0,1|-1/2,7/2))#

So the inverse of the matrix #((7,5),(1,1))# is the matrix #((1/2,-5/2),(-1/2,7/2))#, the same result as before.