# How do you solve the system of equations -x+7y=3 and -4x+28y=12 by the elimination method?

Feb 24, 2017

Infinite solutions.

#### Explanation:

If we multiply $- x + 7 y = 3$ by $4$ it leads to

$- 4 x + 28 y = 12$, the same as the second equation.

Hence $- x + 7 y = 3$ and $- 4 x + 28 y = 12$ represent same line

or two coinciding lines.

Now as $- x + 7 y = 3$ leads to $x = 7 y - 3$, any point $\left(7 k - 3 , k\right)$ falls on these lines and hence the sets of equation has infinite solutions as putting any value of $k$ will give a solution to 'two' equations.

Elimination method

Trying elimination method we have obtained $x = 7 y - 3$ from first equation and putting this in second eliminates $x$ and we get

$- 4 \left(7 y - 3\right) + 28 y = 12$

or $- 4 \times 7 y - 4 \times \left(- 3\right) + 28 y = 12$

or $- 28 y + 12 + 28 y = 12$

or $12 = 12$

When we have such pair of equations, as may be observed, we find that other variable too is eliminated and we get $a = a$, where $a$ is a real number (here $a = 12$). This denotes 'two' coinciding lines (when graphed) and infinite solutions.

Note - if we get two unequal numbers $a = b$, where $a$ and $b$ are two different numbers, this means two parallel lines. For example one can try for $- x + 7 y = 8$ and $- 4 x + 28 y = 12$.