Question

5)
-4s + 2t - 13 = 0
8s - 6t = 42

6)
5y - 20z = 45
y - 4z = 9

Explanation:

For problem 5:

Step1: Rearrange first equation

$-4s + 2t = 13$

Step2: Multiply eq1 by 2

$2(-4s + 2t) = 2\times13$
$\implies -8s + 4t = 26$

Step3: Add to second equation

$(-8s + 4t) + (8s - 6t) = 26 + 42$
$\implies -2t = 68$

Step4: Solve for $t$

$t = \frac{68}{-2} = -34$

Step5: Substitute $t$ into eq1

$-4s + 2(-34) = 13$
$\implies -4s - 68 = 13$
$\implies -4s = 81$
$\implies s = -\frac{81}{4}$

Step1: Multiply second equation by 5

$5(y - 4z) = 5\times9$
$\implies 5y - 20z = 45$

Step2: Compare to first equation

The modified second equation is identical to the first equation $5y - 20z = 45$. This means the two equations represent the same line, so there are infinitely many solutions.

Step3: Express $y$ in terms of $z$

From $y - 4z = 9$, we get $y = 4z + 9$, where $z$ is any real number.

Answer:

$s = -\frac{81}{4}$, $t = -34$

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For problem 6: