Question

1. pq = 3x + 3. qr = 4x + 4. rs = 5x - 15. qs = 100. solve for x. find pq, qr, and rs.

Explanation:

Step1: Analyze the segment addition

We know that \( QS = QR + RS \). Given \( QR = 4x + 4 \), \( RS = 5x - 15 \), and \( QS = 100 \). So we set up the equation:
\( (4x + 4)+(5x - 15)=100 \)

Step2: Simplify the left - hand side

Combine like terms: \( 4x+5x + 4-15 = 100 \), which simplifies to \( 9x-11 = 100 \)

Step3: Solve for x

Add 11 to both sides: \( 9x=100 + 11=111 \)
Then divide both sides by 9: \( x=\frac{111}{9}=\frac{37}{3}\approx12.33 \) (Wait, maybe there is a mistake in the problem understanding. Wait, maybe \( QS=PQ + QR\)? Wait, no, the points are in order P - Q - R - S? Wait, if the order is P, Q, R, S, then \( QS=QR + RS\) and \( PS = PQ+QR + RS\). But maybe the order is P - Q - R - S, and we have \( QS=QR + RS\). But let's re - check. Wait, maybe the problem is that \( QS=PQ + QR\)? No, the given is \( PQ = 3x + 3\), \( QR=4x + 4\), \( RS = 5x-15\), \( QS = 100\). Wait, maybe the correct segment addition is \( QS=QR + RS\). Let's proceed with the equation \( 4x + 4+5x-15 = 100\)
\( 9x-11 = 100\)
\( 9x=111\)
\( x=\frac{111}{9}=\frac{37}{3}\approx12.33\). But this seems odd. Wait, maybe the order is P - Q - R - S, and \( QS=PQ + QR\)? No, \( PQ\) is from P to Q, \( QR\) is from Q to R, so \( PR=PQ + QR\), and \( QS=QR + RS\). Wait, maybe there is a typo, but let's assume the equation \( QS=QR + RS\) is correct.

Wait, maybe I made a mistake. Let's re - do the equation:

\( QR+RS=QS\)

\( (4x + 4)+(5x-15)=100\)

\( 4x+5x+4 - 15=100\)

\( 9x-11 = 100\)

\( 9x=100 + 11=111\)

\( x=\frac{111}{9}=\frac{37}{3}\approx12.33\)

Now, find \( PQ=3x + 3=3\times\frac{37}{3}+3=37 + 3 = 40\)

\( QR=4x + 4=4\times\frac{37}{3}+4=\frac{148}{3}+4=\frac{148 + 12}{3}=\frac{160}{3}\approx53.33\)

\( RS=5x-15=5\times\frac{37}{3}-15=\frac{185}{3}-15=\frac{185 - 45}{3}=\frac{140}{3}\approx46.67\)

Wait, but \( QR+RS=\frac{160}{3}+\frac{140}{3}=\frac{300}{3}=100\), which matches \( QS = 100\). And \( PQ = 40\), \( QR=\frac{160}{3}\), \( RS=\frac{140}{3}\)

But maybe the intended segment addition was \( QS=PQ + QR\)? Let's check that case. If \( QS=PQ + QR\), then:

\( 3x + 3+4x + 4=100\)

\( 7x+7 = 100\)

\( 7x=93\)

\( x=\frac{93}{7}\approx13.29\), but then \( RS = 5x-15=5\times\frac{93}{7}-15=\frac{465}{7}-\frac{105}{7}=\frac{360}{7}\approx51.43\), and there is no relation to \( QS\) in that case. So the first case ( \( QS = QR+RS\)) is correct.

Answer:

\( x=\frac{37}{3}\), \( PQ = 40\), \( QR=\frac{160}{3}\), \( RS=\frac{140}{3}\)