How do you solve the simultaneous equations #y = x^2 + 3x# and #y = 6 - 2x#?

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Answer 1 · 2015-08-03T00:10:53

#color(red)(x=-6,y=18)# and #color(red)(x= 1,y=4)#

One way is to use the method of elimination.

Step 1. Enter the equations.

[1] #y=x^2+3x#
[2] #y=6-2x#

Step 2. Solve for one of the variables in terms of the other.

[2] #y=6-2x#

Since this is already done for us, we can go on to the next step.

Step 3. Substitute Equation 2 in Equation 1 and solve for #x#.

#6-2x=x^2+3x#

#x^2 +5x-6=0#

#(x+6)(x-1)=0#

#x+6=0# and #x-1=0#

#x=-6# and #x-1#

Step4. Substitute each value of #x# in Equation 2

If #x=-6#,
#y=6-2(-6) = 6+12#
#y=18#

If #x=1#,
#y=6-2(1) = 6-2#
#y=4#

Solutions: #x=-6,y=18# and #x= 1,y=4#

Check: Substitute the values of #x# and #y# in Equations 1 and 2.

Check **#1**: (#-6,18#)

#18=(-6)^2+3(-6)#
#18=36-18#
#18=18#

#18=6-2(-6)#
#18 = 6+12#
#18=18#

It checks!

Check **#2**: (#1,4#)

#4 = 1^2 +3(1)#
#4=1+3#
#4=4#

#4=6-2(1)#
#4 = 6-2#
#4=4#

It checks!

Our solutions are correct.