dot product with unknowns of the vectors

Question

$a = (8 , y)$ $b = (2, 3)$ $c = (x, y)$

if $a \cdot c = 10$ and $b \cdot c = 8$, find the values of x and y.

I did up to

$10 = 8x i + y^2 j$

$8 = 2x i + 3y j$

Answer 1

$a\cdot b=8\cdot2+3\cdot y=16+3y$

Now solve: $16+3y=10 \Rightarrow 3y=-6 \Rightarrow \boxed{y=-2}$.

$b\cdot c=2x+3y$

Since we already found $y$ to be $-2$, we now have to solve: $2x-6=8 \Rightarrow \boxed{x=7}$.

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Edit

$a\cdot c=10 \Leftrightarrow 8x+y^2=10$ $(1)$

$b\cdot c=8 \Leftrightarrow 2x+3y=8$ $(2)$

$(2)$ gives $2x=8-3y \Rightarrow x=4-\frac32y$

Plug this into $(1)$ to get $8(4-\frac32y)+y^2=10 \Rightarrow 32-12y+y^2=10 \Rightarrow y^2-12y+22=0 \Rightarrow y=6\pm\sqrt{14}$

Plug this back into $(1)$ to get: $8x+(6\pm\sqrt{14})^2=10 \Rightarrow 8x+36\pm12\sqrt{14}+14=10 \Rightarrow 8x=10-14-36\pm12\sqrt{40} \Rightarrow x=-5\pm\frac32\sqrt{14}$

The pairs $(x=-5\pm\frac32\sqrt{14},y=6\mp\sqrt{14})$ represent the accepted values for $x$ and $y$.

Answer 2

You just have to solve the system:

$$\begin{cases}a\cdot b=10\\b\cdot c=8\end{cases}\implies \begin{cases}16+3y=10\\2x+3y=8\end{cases}\implies ...$$