How do you solve the following linear system: #2x+y=-3/2, 6x+3y=5#?

2 Answers
Mar 13, 2018

Answer:

There is no solution for the pair of equations.

Explanation:

#2x+y=-3/2#

Multiply 3 both sides

#=> 3(2x+y)=3xx-3/2#

#=> 6x+3y = -9/2#

And , #6x+3y=5#

You see the LHS of both the equations are equal but the RHS of both the equations are unequal.

Thus , no solution exists for the given pair of linear equations.

Mar 13, 2018

Answer:

#"no solution"#

Explanation:

#2x+y=-3/2to(1)#

#6x+3y=5to(2)#

#"From equation "(1)" we obtain"#

#y=-3/2-2x#

#color(blue)"Substitute "y=-3/2-2x" into equation "(2)#

#6x+3(-3/2-2x)=5#

#rArrcancel(6x)-9/2cancel(-6x)=5#

#rArr-9/2=5#

#"Obviously this is not a true statement hence no solution"#

#"Consider the equations in "color(blue)"slope-intercept form"#

#(1)toy=-3/2-2x#

#(2)toy=-2x+5/3#

#"both lines have "m=-2rArr" parallel lines"#

#"thus they never intersect and so have no solution"#
graph{(y+2x+3/2)(y+2x-5/3)=0 [-10, 10, -5, 5]}