How do you solve the following systems of equations algebraically?

Interesting! If you think of the graphs in each case the first equation is a circle and the second is a straight line. If I am not mistaken the lines are both tangents to their circles.

I'm going to do these by finding an expression for y^2 in terms of x and so producing an equation in x only. It will be a quadratic equation so you might expect two solutions but if it is a tangent line then there will only be one answer.

1, -3x + 4y = 25

4y = 3x + 25

16y^2 = (3x + 25)^2 = 9x^2 + 150x + 625

x^2 + y^2 = 25

y^2 = 25 - x^2

16y^2 = 400 - 16x^2

9x^2 + 150x + 625 = 400 - 16x^2

25x^2 +150x 225 = 0

x^2 + 6x + 9 = 0

(x+3)^2 = 0

x = -3 and substituting in one of the original equations y = 4

2. 3y = 10 - x

9y^2 = 100 - 20x + x^2

9y^2 = 90 - 9x^2

100 - 20x + x^2 = 90 - 9x^2

10x^2 - 20x + 10 = 0

x^2 - 2x + 1 = 0

(x - 1)^2 = 0

x = 1 and y = 3

First one has something wrong.

x^(2)+y^(2)=10_x+3y=10

Since 3y does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 3y from both sides.

x^(2)+y^(2)=10_x=-3y+10

Replace all occurrences of x with the solution found by solving the last equation for x. In this case, the value substituted is -3y+10.

(-3y+10)^(2)+y^(2)=10_x=-3y+10

Squaring an expression is the same as multiplying the expression by itself 2 times.

(-3y+10)(-3y+10)+y^(2)=10_x=-3y+10

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.

(-3y*-3y-3y*10+10*-3y+10*10)+y^(2)=10_x=-3y+10

Simplify the FOIL expression by multiplying and combining all like terms.

(9y^(2)-60y+100)+y^(2)=10_x=-3y+10

Since 9y^(2) and y^(2) are like terms, add y^(2) to 9y^(2) to get 10y^(2).

10y^(2)-60y+100=10_x=-3y+10

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

10y^(2)-60y+90=0_x=-3y+10

Factor out the GCF of 10 from each term in the polynomial.

10(y^(2))+10(-6y)+10(9)=0_x=-3y+10

Factor out the GCF of 10 from 10y^(2)-60y+90.

10(y^(2)-6y+9)=0_x=-3y+10

For a polynomial of the form x^(2)+bx+c, find two factors of c (9) that add up to b (-6). In this problem -3*-3=9 and -3-3=-6, so insert -3 as the right hand term of one factor and -3 as the right-hand term of the other factor.

10(y-3)(y-3)=0_x=-3y+10

Combine the two common factors of (y-3) by adding the exponents.

10(y-3)^(2)=0_x=-3y+10

Divide both sides of the equation by 10. Dividing 0 by any non-zero number is 0.

(y-3)^(2)=0_x=-3y+10

Set each of the factors of the left-hand side of the equation equal to 0.

y-3=0_x=-3y+10

Since -3 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 3 to both sides.

y=3_x=-3y+10