What are the intercepts of #-7x+4y=13#?

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Answer 1 · 2016-01-07T10:58:29

#x# intercept:

#(-13/7;0)#

#y# intercept:

#(0;13/4)#

you have to write the expression in the form:

#y=f(x)#

#4y=7x+13#

#y=7/4*x+13/4#

#x# intercept:

#y=0#

then:

#7/4*x+13/4=0#

#7/4*x=-13/4#

#7x=-13#

#x=-13/7#

then the #x# intercept point is

#(-13/7;0)#

#y# intercept:

#x=0#

#y=7/4*0+13/4#

#y=0+13/4=13/4#

Then the #y# intercept point is:

#(0;13/4)#