How do you solve the following system?: #y=2x-3 , 5x + 3y = -3 #

Question

How do you solve the following system?: #y=2x-3 ,    5x + 3y = -3   #

Savannah Anderson · Accepted Answer

#x,y=6/11,-21/11#

#color(blue)(y=2x-3#

#color(blue)(5x+3y=-3#

Substitute the value of #y# to the second equation

#rarr5x+3(2x-3)=-3#

Use distributive property #color(brown)(a(b+c)=ab+ac#

#rarr5x+6x-9=-3#

#rarr11x-9=-3#

#rarr11x=-3+9#

#rarr11x=6#

#color(green)(rArrx=6/1

Substitute the value of #x# to the first equation

#rarry=2(6/11)-3#

#rarry=12/11-3#

#color(green)(rArry=-21/11#

Jace Anderson · Answer

undefined To solve the system of equations $ y = 2x - 3 $ and $ 5x + 3y = -3 $, you can use the substitution method or the elimination method.

First, solve the first equation for $ y $ in terms of $ x $: $ y = 2x - 3 $.

Then, substitute $ 2x - 3 $ for $ y $ in the second equation: $ 5x + 3(2x - 3) = -3 $.

Now, solve for $ x $:

$ 5x + 6x - 9 = -3 $

$ 11x - 9 = -3 $

$ 11x = 6 $

$ x = \frac{6}{11} $

Next, substitute the value of $ x $ back into either of the original equations to solve for $ y $. Using the first equation is usually simpler:

$ y = 2(\frac{6}{11}) - 3 $

$ y = \frac{12}{11} - 3 $

$ y = \frac{12}{11} - \frac{33}{11} $

$ y = \frac{-21}{11} $

So, the solution to the system of equations is $ x = \frac{6}{11} $ and $ y = \frac{-21}{11} $.