How do you solve the system of equations #-7x + y = - 1# and #2x - 5y = 5#?

3 Answers
smendyka
Jan 30, 2018

See a solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#-7x + y = -1#

#color(red)(7x) - 7x + y = color(red)(7x) - 1#

#0 + y = 7x - 1#

#y = 7x - 1#

Step 2) Substitute #(7x - 1)# for #y# in the second equation and solve for #y#:

#2x - 5y = 5# becomes:

#2x - 5(7x - 1) = 5#

#2x - (5 xx 7x) - (5 xx -1) = 5#

#2x - 35x - (-5) = 5#

#2x - 35x + 5 = 5#

#(2 - 35)x + 5 = 5#

#-33x + 5 = 5#

#-33x + 5 - color(red)(5) = 5 - color(red)(5)#

#-33x + 0 = 0#

#-33x = 0#

#(-33x)/color(red)(-33) = 0/color(red)(-33)#

#(color(red)(cancel(color(black)(-33)))x)/cancel(color(red)(-33)) = 0#

#x = 0#

Step 3) Substitute #0# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 7x - 1# becomes:

#y = (7 xx 0) - 1#

#y = 0 - 1#

#y = -1#

The Solution Is:

#x = 0# and #y = -1#

Or

#(0, -1)#

Jim G.
Jan 30, 2018

#(x,y)to(0,-1)#

Explanation:

#-7x+y=-1to(1)#

#2x-5y=5to(2)#

#"from equation "(1)" we can express y in terms of x"#

#(1)toy=-1+7xto(3)#

#"substitute "y=-1+7x" in equation "(2)#

#2x-5(-1+7x)=5#

#rArr2x+5-35x=5#

#rArr-33x+5=5#

#"subtract 5 from both sides"#

#-33xcancel(+5)cancel(-5)=5-5#

#rArr-33x=0rArrx=0#

#"substitute "x=0" in equation "(3)#

#rArry=-1+0=-1#

#"the point of intersection "=(0,-1)# graph{(y+1-7x)(y-2/5x+1)=0 [-10, 10, -5, 5]}

Cem Sentin
Jan 30, 2018

#x=0# and #y=-1#

Explanation:

#5*(-7x+y)+2x-5y=(-1)*5+5#

#-35x+5y+2x-5y=-5+5#

#-33x=0#, so #x=0#

Thus,

#(-7)*0+y=-1# or #y=-1#

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