3 years ago

SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION. 1. 2x+3y=1 -3x+y=15

1 answer:

6 0

$2x+3y=1 \\
-3x+y=15 \\ \\
\hbox{from the 2nd equation:} \\
-3x+y=15 \\
y=15+3x \\ \\
\hbox{substitute 15+3x for y in the 1st equation:} \\
2x+3(15+3x)=1 \\
2x+45+9x=1 \\
9x+2x=1-45 \\
11x=-44 \\
x=-4 \\ \\
\hbox{solve one of the equations for y}: \\
2x+3y=1 \\
2 \cdot (-4) + 3y=1 \\
-8+3y=1 \\
3y=1+8 \\
3y=9 \\
y=3 \\ \\
\hbox{the answer:} \\
x=-4 \\
y=3$

You might be interested in

Find the value of c that completes the square for x2−15x+c

meriva

Answer:

The value of c is $\frac{225}{4}$ .

Step-by-step explanation:

The perfect square of the difference between two numbers is:

$(x-y)^{2}=x^{2}-2xy+y^{2}$

The expression provided is:

$x^{2}-15x+c$

The expression is a perfect square of the difference between two numbers.

One of the number is x and the other is √c.

Use the above relation to compute the value of c as follows:

$x^{2}-15x+c=(x-\sqrt{c})^{2}\\\\x^{2}-15x+c=x^{2}-2\cdot x\cdot\sqrt{c}+c\\\\15x=2\cdot x\cdot\sqrt{c}\\\\15=2\cdot\sqrt{c}\\\\\sqrt{c}=\frac{15}{2}\\\\c=[\frac{15}{2}]^{2}\\\\c=\frac{225}{4}$