Consider the following system of equations.
Equation 1 | $3x+7y=-6$3x+7y=−6 |
Equation 2 | $2x-y=-17$2x−y=−17 |
Suppose we want to solve this system by using the elimination method and eliminating $y$y.
What value can we multiply Equation 2 by so that the coefficients of $y$y in each equation are opposite numbers?
What equation do we get when we multiply the second equation by $7$7?
Consider this system of equations.
Equation 1 | $\frac{2x}{5}+\frac{3y}{5}=-\frac{7}{5}$2x5+3y5=−75 |
Equation 2 |
$-\frac{1}{4}\left(-5x+\frac{7y}{9}\right)=2$−14(−5x+7y9)=2 |
Consider the following system of equations.
$-8x$−8x | $-$− | $y$y | $=$= | $0$0 |
$-5x$−5x | $+$+ | $3y$3y | $=$= | $6$6 |
We are solving this system using the elimination method.
When we solve a system of equation using the addition method, what happens when the system has no solutions?