How do you solve the system of equations 3x + y = 16 and 2x + y = 11?

2 Answers
Oct 22, 2016

x=5
y=1

Explanation:

Solving simultaneously :
color(lightblue)((1) color(lightblue)rarr 3x+y=16
color(lightgreen)((2) color(lightgreen)rarr 2x+y=11

Multiply color(lightgreen)((2) by -1:
color(white)(xxxxx)3xcancel(+y)=+16
color(white)(xxx)-2xcancel(-y)=-11

Add color(lightblue)((1) & color(lightgreen)((2):
color(white)(xxxxx)color(red)(x=5)

*Substitute the value of color(red)x in color(lightgreen)((2):
2(5)+y=11

color(red)(y=1)

Oct 22, 2016

x = 5 and y=1

Explanation:

The methods shown by other contributors are the elimination method and the graphical method.

Another option is the method of equating variables.

This is particularly useful for finding the point of intersection of two straight lines algebraically.

In the two equations, notice that they have the SAME y-TERM

3x color(red)(+y) = 16" and "2x color(red)(+y) = 11

Make color(red)(y) the subject in each case by re-arranging the equations.

color(red)(y) = 16-3x " and "color(red)(y) = 11-2x

We know that the value of " "color(red)(y)" " will be the same in each equation.

color(white)(xxxxxxxxxxxx)color(red)(y = y)

color(white)(xxx xxx):.11-2x = 16-3x" "larr only x terms.

color(white)(xxx xxx):.3x-2x = 16-11" "larr re-arrange terms

color(white)(xxxxxxxxxx):.x = 5

There are now 2 equations to use to find y

Substitute x = 5 into both to check that you get the same answer.

y= 16-3(5) = 16-15 =1

y = 11-2(5) = 11-10 = 1
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Another option:

Often we forget how useful it is to just look at equations first and we set off doing calculations instead of thinking first.

In 3x +y = 16" and " 2x+y = 11,

the left hand sides only differ by 1x
the right hand sides differ by 5.

x must be be 5.

If you know the value of x you can work out y by inspection.