Solve the system of equations using the substitution method. I can't understand the 5/6x!!
y = 2 + 5/6 * x 4x - 3y = 3
+30 \(\mathrm{Subsititute\:}y=2+\frac{5}{6}\cdot \:x \)
\(\begin{bmatrix}4x-3\left(2+\frac{5}{6}x\right)=3\end{bmatrix}\)
\(\mathrm{Isolate}\:x\:\mathrm{for}\:4x-3\left(2+\frac{5}{6}x\right)=3:\quad x=6\)
\(\mathrm{For\:}y=2+\frac{5}{6}\cdot \:x\)
\(\mathrm{Subsititute\:}x=6\)
\(2+\frac{5}{6}\cdot \:6=7\)
\(y=7, \text{Solution is x=6, y=7}\)
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+37146 y = 2 + 20/6 x - 3y add 3 y to both sides of the equation
4y = 2 + (20/6) x your original Q says y = 3 substitute that in
12 = 2 + 20/6 x subtract 2 from both sides
10 = 20/6 x multiply both sides by 6/20
60/20 = x = 3
Now I see your question has TWO Equations....please put some spaces in there !
y = 2 + 5/6 * x
4x - 3y = 3 Is much more clear.....see whoisjoe answer below....
+30 Hmm. We seem to be disagreeing. I think you forgot that y = 2 + 5/6x is a different term than 4x-3y=3, lol. Guest probably should have put something to tell us that they are 2 different terms, unless I am the mistaken one. Oh I see you didn't see that it was a system of equations so you just solved for x!
