Solve these simultaneous equations. $2x-3y=19$ $3x+2y=-4$ Answer
x = ... y = ... It is given that $4^p = 5$, $5^q = 6$, $6^r = 7$ and $7^s = 8$. Find the exact value of $pqrs$. Answer $pqrs =$ ...

1. Make the coefficients of $y$ equal in magnitude.

$2x-3y=19 \quad \times 2 \Rightarrow 4x-6y=38$
$3x+2y=-4 \quad \times 3 \Rightarrow 9x+6y=-12$

2. Eliminate $y$ and solve for $x$.

Add the two equations:
$4x-6y+9x+6y=38+(-12)$
$13x=26$
$x=2$

3. Substitute $x=2$ into one original equation to find $y$.

Using $3x+2y=-4$,
$3(2)+2y=-4$
$6+2y=-4$
$2y=-10$
$y=-5$

1. Rewrite each given equation as a power expression for the next number in the chain.

From the given equations,
$5=4^p,$
$6=5^q,$
$7=6^r,$
$8=7^s.$

2. Substitute repeatedly to express $8$ as a power of $4$.

$8=7^s=(6^r)^s=6^{rs}$
$=(5^q)^{rs}=5^{qrs}$
$=(4^p)^{qrs}=4^{pqrs}$
So,
$8=4^{pqrs}.$

3. Compare powers of $2$ to find $pqrs$.

Since
$8=2^3$
and
$4^{pqrs}=(2^2)^{pqrs}=2^{2pqrs},$
we have
$2^{2pqrs}=2^3.$
Hence,
$2pqrs=3$
$pqrs=\frac{3}{2}.$