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Zolol [24]
3 years ago
10

Look at the system of equations below.

Mathematics
1 answer:
Paul [167]3 years ago
7 0

Answer:

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=2,\:y=-1

Step-by-step explanation:

Considering the system of the equation

\begin{bmatrix}y=-2x+3\\ 4x-3y=11\end{bmatrix}

\mathrm{Subsititute\:}y=-2x+3

\begin{bmatrix}4x-3\left(-2x+3\right)=11\end{bmatrix}

\mathrm{Isolate}\:x\:\mathrm{for}\:4x-3\left(-2x+3\right)=11

4x-3\left(-2x+3\right)=11

4x+6x-9=11

10x-9=11

10x-9+9=11+9

10x=20

x=2

\mathrm{For\:}y=-2x+3

\mathrm{Subsititute\:}x=2

y=-2\cdot \:2+3

y=-1

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=2,\:y=-1

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Answer:

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Step-by-step explanation:

Hello

To find Sn, we need to draw out equations for each a₇ and a₁₉

In an arithmetic progression,

Sn = a + (n-1)d

Where Sn = sum of the A.P

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Put equation (iii) into equation (ii)

140 = (32 - 6d) + 18d

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