How do you solve this system of equations using substitution 3x-2y=4 4x-3y=7

(2) Answers

[latex]\begin{cases} 3x-2y=4 \\ 4x-3y=7 \end{cases} \\ \begin{cases} -2y=4-3x \\ 4x-3y=7\end{cases} \\ \begin{cases} y= \frac{4-3x}{-2} \\ 4x-3y=7 \end{cases} \\ \begin{cases}y=\frac{4-3x}{-2}\\ 4x-3(\frac{4-3x}{-2})=7\end{cases} \\ \begin{cases}4x+6-4,5x=7\\y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases}6- \frac{1}{2}x=7\\ y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases} - \frac{1}{2}x=1\\y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases}x=-2 \\y=\frac{4-3x}{-2} \end{cases} [/latex] [latex]\begin{cases}x=-2 \\ y= \frac{4-3(-2)}{-2} \end{cases} \\ \begin{cases}x=-2 \\ y=-5 \end{cases}[/latex]


so you try to get one of the terms to cancel so I decide to cancel the 'y' terms 3x-2y=4 4x-3y=7 we can find the least common multiply of the y terms (3 and 2) which is 6 so 3x-2y=4 or 9x-6y=12 4x-3y=7 or 8x-6y=14 make the first equation negative and add to the second    -9x+6y=-12 +  8x - 6y=14 the y terms cancel and we get -x=2 so x=-2 subsitute x=-2 into any equation (i choose first) 3x-2y=4 3(-2)-2y=4 -6-2y=4 add 6 to both sides -2y=10 multiply both sides by 1/2 -y=5 multiply both sides by -1 y=-5 x=-2 and y=-5

Add answer