What are x and y if #y = 4x + 3# and #2x + 3y = -5#?

show below

#y = 4x + 3#..........1

#2x + 3y = -5#..........2

put 1 in 2

#2x + 3(4x + 3)= -5#

#2x+12x+9=-5#

#14x=-14#

#x=-1#

#y = 4(-1) + 3=-4+3=-1#

There are two ways to algebraically solve for #x# and #y#.

Method 1: Substitution

Through this method, we solve to a variable in one equation and plug it in to the other. In this case, we already know the value of #y# in the first equation. Therefore, we can substitute it it for #y# in the second equation and solve for #x#.

#y=4x+3#
#2x+3(4x+3)=-5#
#2x+12x+9=-5#
#14x=-14#
#x=-1#

Now, we just need to plug #x# back in to one of the equations to solve for #y#. We can use the first equation because #y# is already isolated, but both will yield the same answer.

#y=4(-1)+3)#
#y=-4+3#
#y=-1#

Therefore, #x# is #-1# and #y# is #-1#.

Method 2: Elimination

Through this method, the equations are subtracted so that one of the variables is eliminated. To do this, we must isolate the constant number. In other words, we put #x# and #y# on the same side, like in the second equation.

#y=4x+3#
#0=4x-y+3#
#-3=4x-y#

Now, the equations are both in the same form. However, to eliminate one of the variables, we must get #0# when the equations are subtracted. This means we must have the same coefficients on the variable. For this example, let's solve for #x#. In the first equation, #x# has a coefficient of #4#. Thus, we need #x# in the second equation to have the same coefficient. Because #4# is #2# times its current coefficient of #2#, we need to multiply the entire equation by #2# so it stays equivalent.

#2(2x+3y)=2(-5)#
#4x+6y=-10#

Next, we can subtract the two equations.

#4x+6y=-10#
#-(4x-y=-3)#
–––––––––––––––––––
#0x+7y=-7#

#7y=-7#
#y=-1#

As with the first method, we plug this value back in to find #x#.

#-1=4x+3#
#-4=4x#
#-1=x#

Therefore, #x# is #-1# and #y# is #-1#.