Solve the system of equation using substitution method

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Here are the equations:

$$c+6d=7\text{ and }-c-2d=-2$$

I tried to do it but in the end, I get the wrong answer. The intersection point is a decimal.

This is how I tried doing it:

I first simplified the first equation which would be $$d=-\frac{1}{6}c+\frac{7}{6}$$

Then I substituted and this is how it looks: $$-c-2\left(-\frac{1}{6}c+\frac{7}{6}\right)=-2$$

I simplified the equation and got $$c=\frac{13}{6}-\frac{6}{3}$$

for the answer, I got $c=\frac{1}{6}$

last I substituted $-c-2d=-2$ to $-\frac{1}{6}-2d=-2$

I got $d=\frac{11}{12}$ but I got it wrong

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For advice, it is often preferred to avoid fractions whenever possible. Rather than solving for $d$ and using that to substitute, it would have been cleaner to solve for $c$ here, it would have just been $c=7-6d$ with no fractions necessary.

Following your work, you got $d=\frac{7}{6}-\frac{1}{6}c$. This is correct

You continued with $-c-2d=-2$ and substituting in giving $-c-2\left(\frac{7}{6}-\frac{1}{6}c\right) = -2$ which is good.

You say that you "simplified the equation and got..." $c=\frac{13}{6}-\frac{6}{3}$. You do not show how you simplified to get to this, but this is wrong. Let us work through the simplification in more detail:

$$-c-2\left(\frac{7}{6}-\frac{1}{6}c\right)=-2$$

Distribute the negative 2 on the parentheses

$$-c+(-2)\times\frac{7}{6} + (-2)\times(-\frac{1}{6}c)=-2$$

Note that $6=2\times 3$ and so $(-2)\times\frac{x}{6}=(-2)\times\frac{x}{2\times 3}=-\frac{x}{3}$ so this simplifies to:

$$-c-\frac{7}{3}+\frac{1}{3}c = -2$$

move constants to one side and combine

$$-\frac{2}{3}c = \frac{1}{3}$$

multiply both sides by $-\frac{3}{2}$

$$c = -\frac{1}{2}$$

From here we can finish the problem by back-substitution.


If we were to have started by solving for $c$ we would have had $c=7-6d$, and then substituting in to the second expression we would have had $(7-6d)+2d=2$ which is easier to work with than having to manage fractions at the start.

If we were to continue with this,

$$(7-6d)+2d=2$$

$$7-4d=2$$

$$4d=5$$

$$d=\frac{5}{4}$$

Then back substituting:

$$c+2d=2$$

$$c+2\left(\frac{5}{4}\right)=2$$

$$c+\frac{5}{2}=2$$

$$c = -\frac{1}{2}$$