Let p = (x + 4y),a + (2x + y + 1),b and q = ( | vedclass

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

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$x = 2, y = -1$

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(B) Given $p = (x + 4y)\,a + (2x + y + 1)\,b$ and $q = (y - 2x + 2)\,a + (2x - 3y - 1)\,b$.Since $3p = 2q$,we have:$3[(x + 4y)\,a + (2x + y + 1)\,b] = 2[(y - 2x + 2)\,a + (2x - 3y - 1)\,b]$$(3x + 12y)\,a + (6x + 3y + 3)\,b = (2y - 4x + 4)\,a + (4x - 6y - 2)\,b$Since $a$ and $b$ are non-collinear,we compare the coefficients:$3x + 12y = 2y - 4x + 4 \implies 7x + 10y = 4 \quad (i)$$6x + 3y + 3 = 4x - 6y - 2 \implies 2x + 9y = -5 \quad (ii)$Multiplying $(i)$ by $2$ and $(ii)$ by $7$:$14x + 20y = 8 \quad (iii)$$14x + 63y = -35 \quad (iv)$Subtracting $(iii)$ from $(iv)$:$43y = -43 \implies y = -1$Substituting $y = -1$ into $(ii)$:$2x + 9(-1) = -5 \implies 2x - 9 = -5 \implies 2x = 4 \implies x = 2$Thus,$x = 2$ and $y = -1$.

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

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