How do you solve 35x - 5y = 20 and y = 7x + 4 using substitution?

Aug 30, 2017

See a solution process below:

Explanation:

Step 1) Because the second equation is already solved for $y$ we can substitute $\left(7 x + 4\right)$ for $y$ in the first equation and solve for $x$:

$35 x - 5 y = 20$ becomes:

$35 x - 5 \left(7 x + 4\right) = 20$

$35 x - \left(5 \cdot 7 x\right) - \left(5 \cdot 4\right) = 20$

$35 x - 35 x - 20 = 20$

$35 x - 35 x - 20 = 20$

$0 - 20 = 20$

$- 20 \ne 20$

This indicates there are no solutions to this system of equations. Or, no solution set is the empty or null set: $\left\{\emptyset\right\}$

This also indicates, if there is no solutions, that these equations represent parallel lines.

Aug 30, 2017

"Substitute" the expression for y in the second equation into the first one.

Explanation:

35x − 5(7x + 4) = 20
35x − 35x - 20 = 20
$0 = 40$
0 does not = 40, so there is no solution to this set.

Converting both to slope-intercept form gives us:
35x − 5y = 20 ; $- 5 y = - 35 x + 20$ Dividing by -5 we get:
$y = 7 x - 4$; compared to the second equation
$y = 7 x + 4$

They are parallel lines (same slope) with different intercepts (4 and -4).