Find the x- and y-intercepts of5x-3y = 12.x-intercept:y-intercept:

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

8 0

Answer:

-5x -5x

-3y=-5x +12

/-5x /-5x

-5-3y=x+12

/-3 /-3

-5+y=x-4

+5 +5

y=x+1<---(y intercept(and it is its slop))

5x-3y = 12

/-3y /-3y

5x=-3y+12

/5 /5

x=-3y+2.4

/-3 /-3

-3+x=y+2.4

+3 +3

x=y+5.4<---(x intercept(not the slop because the y is not the first number))

You might be interested in

Three numbers X, Y, and Z are in the ratio 2:7:8. If 12 is subtracted from Y, then three numbers form a geometric sequence (in t

Dimas [21]

A geometric sequence is a sequence in which there is a common ratio between any two consecutive terms. In this case if X:Y:Z are in the ratio of 2:7:8 the multiplying by a constant k, we have X=2k, Y= 7k and Z=8k.
Then if X, Y-12, Z form a Geometric sequence, it means X/Y-12=Y-12/Z which is the same as 2k/7k-12=7k-12/8k if we cross multply, we get
16k²= 49k²-168k +144
33k²-168k+144 =0 solving for k
k = 4 or 1.091 if we take the whole number to find the values of X,Y and z,
X= 8, Y= 28 and Z=32

3 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$