(a) Solve the simultaneous equation : \(6y + 5x =... | WAEC SSCE - General Mathematics - 1996 (Essay)

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(a) Solve the simultaneous equation : \(6y + 5x = 12 ; 4y - 3x = 11\).

(b)

In the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y cm. Find the value of y.

Answer

Answer Details

(a) Solving the simultaneous equations

Label the equations:

\[5x + 6y = 12 \quad (1)\]\[-3x + 4y = 11 \quad (2)\]

Eliminate \(x\). Multiply (1) by 3 and (2) by 5 so the \(x\)-terms cancel:

\[15x + 18y = 36 \quad (3)\]\[-15x + 20y = 55 \quad (4)\]

Add (3) and (4):

\[38y = 91 \;\Rightarrow\; y = \frac{91}{38} = 2\tfrac{15}{38} \approx 2.39\]

Find \(x\). Multiply (1) by 2 and (2) by 3:

\[10x + 12y = 24, \qquad -9x + 12y = 33\]

Subtract the second from the first:

\[19x = 24 - 33 = -9 \;\Rightarrow\; x = -\frac{9}{19} \approx -0.47\]

Check in (2): \(-3\left(-\tfrac{9}{19}\right) + 4\left(\tfrac{91}{38}\right) = \tfrac{54}{38} + \tfrac{364}{38} = \tfrac{418}{38} = 11.\) Correct.

Hence \(x = -\dfrac{9}{19}\), \(y = \dfrac{91}{38}\).

(b) Finding \(y = |AD|\)

From the diagram, \(ADC\) is a straight line with \(BD \perp AC\), so \(BD\) is the altitude from the right angle \(B\) to the hypotenuse \(AC\) of the right-angled triangle \(ABC\) (\(\angle ABC = 90^\circ\)). Here \(|BD| = 36\text{ cm}\), \(|DC| = 48\text{ cm}\) and \(|AD| = y\).

For a right-angled triangle, the altitude to the hypotenuse is the geometric mean of the two segments it makes on the hypotenuse:

\[|BD|^{2} = |AD|\times|DC|\]\[36^{2} = y \times 48\]\[1296 = 48y\]\[y = \frac{1296}{48} = 27\]

Hence \(y = 27\text{ cm}\).

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(a) Solve the simultaneous equation : \(6y + 5x = 12 ; 4y - 3x = 11\). (b) In the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y c...

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(a) Solve the simultaneous equation : \(6y + 5x = 12 ; 4y - 3x = 11\). (b) In the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y c...

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