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

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Answer 1 · 2016-01-07T10:58:29

$x$ $x$ intercept:

$(- \frac{13}{7}; 0)$ $(-13/7;0)$

$y$ $y$ intercept:

$(0; \frac{13}{4})$ $(0;13/4)$

you have to write the expression in the form:

$y = f (x)$ $y=f(x)$

$4 y = 7 x + 13$ $4y=7x+13$

$y = \frac{7}{4} \cdot x + \frac{13}{4}$ $y=7/4*x+13/4$

$x$ $x$ intercept:

$y = 0$ $y=0$

then:

$\frac{7}{4} \cdot x + \frac{13}{4} = 0$ $7/4*x+13/4=0$

$\frac{7}{4} \cdot x = - \frac{13}{4}$ $7/4*x=-13/4$

$7 x = - 13$ $7x=-13$

$x = - \frac{13}{7}$ $x=-13/7$

then the $x$ $x$ intercept point is

$(- \frac{13}{7}; 0)$ $(-13/7;0)$

$y$ $y$ intercept:

$x = 0$ $x=0$

$y = \frac{7}{4} \cdot 0 + \frac{13}{4}$ $y=7/4*0+13/4$

$y = 0 + \frac{13}{4} = \frac{13}{4}$ $y=0+13/4=13/4$

Then the $y$ $y$ intercept point is:

$(0; \frac{13}{4})$ $(0;13/4)$

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