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3 years ago

How many solutions does this system have? y = 3 x minus 5. y = negative x + 4. one two an infinite number no solution

Mathematics

2 answers:

____ [38]3 years ago

8 0

Answer:

Step-by-step explanation:

mote1985 [20]3 years ago

6 0

Answer:

One

Step-by-step explanation:

y = 3 x - 5.

y = -x + 4.

x=9/4 and y=7/4

1 solution (9/4 , 7/4)

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Jane’s buying clothes. She can get 5 shirts & 4 sweaters for $229 or she can get 2 shirts & 3 sweaters for $128. How muc

makkiz [27]

Cost of 1 shirt is $\$25$ and cost of one sweater is $\$26$

Solution:

Given that Jane’s buying clothes.

She can get 5 shirts & 4 sweaters for $\$229$

She can get 2 shirts & 3 sweaters for $\$128$

Need to determine cost of 1 shirt and 1 sweater.

Let assume cost of 1 shirt = $\$x$

And assume cost of 1 sweater = $\$y$

Let us create equations from give conditions

First condition is Cost of 5 shirt + cost of 4 sweater = $\$229$

$\begin{array}{l}{\text { cost of 5 shirts }=5 \times \text { cost of 1 shirt }=5 \times x=5 x} \\\\ {\text { cost of 4 sweater }=4 \times \text { cost of 1 sweater }=4 \times y=4 y}\end{array}$

So using first condition equation which we get is as follows

$5 x+4 y=229 \rightarrow (1)$

Second condition is Cost of 5 shirt + cost of 4 sweater = $\$128$

$\begin{array}{c}{\text { cost of } 2 \text { shirts }=2 \times \text { cost of } 1 \text { shirt }=2 \times x=2 x} \\\\ {\text { cost of } 3 \text { sweater }=3 \times \text { cost of } 1 \text { sweater }=3 \times y=3 y}\end{array}$

So using second condition equation which we get is as follows

$2 \mathrm{x}+3 \mathrm{y}=128\quad\rightarrow (2)$

Now we have two equations to be solved

$\begin{array}{l}{5 x+4 y=229 \rightarrow (1)} \\\\ {2 x+3 y=128 \rightarrow (2)}\end{array}$

On multiplying equation (1) by 2 and equation (2) by 5 to make coefficients of x equal in both equations, we get

$\begin{array}{l}{2 x(5 x+4 y)=2\times 229} \\\\ {10 x+8 y=458 \rightarrow (3)} \\\\ {5 x(2 x+3 y)=5 \times 128} \\\\ {10 x+15 y=640 \rightarrow (4)}\end{array}$

On subtracting (3) from (4), we get

$\begin{array}{l}{(10 \mathrm{x}-10 \mathrm{x})+(15 \mathrm{y}-8 \mathrm{y})=640-458} \\\\ {\Rightarrow 7 \mathrm{y}=182} \\\\ {\Rightarrow \mathrm{y}=\frac{182}{7}=26}\end{array}$

On substituting y = 26 in equation (1) we get

$\begin{array}{l}{\Rightarrow 5 \mathrm{x}+(4 \times 26)=229} \\\\ {\Rightarrow 5 \mathrm{x}=229-104} \\\\ {\Rightarrow 5 \mathrm{x}=125} \\\\ {\Rightarrow \mathrm{x}=\frac{125}{5}=25}\end{array}$