Let p = (x + 4y)vec{a} + (2x + y + 1)vec{b} a | vedclass

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(B) Given $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$.Since $3p = 2q$,we have:$3[(x + 4y)\vec{a}

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$x = 2, y = -1$

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(B) Given $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$.Since $3p = 2q$,we have:$3[(x + 4y)\vec{a} + (2x + y + 1)\vec{b}] = 2[(y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}]$$(3x + 12y)\vec{a} + (6x + 3y + 3)\vec{b} = (2y - 4x + 4)\vec{a} + (4x - 6y - 2)\vec{b}$Since $\vec{a}$ and $\vec{b}$ are non-collinear,we can equate the coefficients:$1) 3x + 12y = 2y - 4x + 4 \Rightarrow 7x + 10y = 4$$2) 6x + 3y + 3 = 4x - 6y - 2 \Rightarrow 2x + 9y = -5$Multiplying equation $(1)$ by $2$ and equation $(2)$ by $7$:$14x + 20y = 8$$14x + 63y = -35$Subtracting the equations: $(63y - 20y) = -35 - 8 \Rightarrow 43y = -43 \Rightarrow y = -1$Substituting $y = -1$ into $2x + 9y = -5$:$2x + 9(-1) = -5 \Rightarrow 2x - 9 = -5 \Rightarrow 2x = 4 \Rightarrow x = 2$Thus,$x = 2$ and $y = -1$.

Let $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$,where $\vec{a}$ and $\vec{b}$ are non-collinear vectors. If $3p = 2q$,then the values of $x$ and $y$ are:

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