How do you solve the following linear system: 2x+y=-3/2, 6x+3y=52x+y=−32,6x+3y=5?
2 Answers
There is no solution for the pair of equations.
Explanation:
Multiply 3 both sides
And ,
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.
Explanation:
2x+y=-3/2to(1)2x+y=−32→(1)
6x+3y=5to(2)6x+3y=5→(2)
"From equation "(1)" we obtain"From equation (1) we obtain
y=-3/2-2xy=−32−2x
color(blue)"Substitute "y=-3/2-2x" into equation "(2)Substitute y=−32−2x into equation (2)
6x+3(-3/2-2x)=56x+3(−32−2x)=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]}
