We have a system of simultaneous equations: a linear equation 3x plus 2y equals 13, and a circle equation x squared plus y squared equals 25. The linear equation represents a straight line, while the circle equation represents a circle with radius 5 centered at the origin. The solutions are the intersection points of these two curves.
First, we solve the linear equation for y. Starting with 3x plus 2y equals 13, we subtract 3x from both sides to get 2y equals 13 minus 3x. Then we divide by 2 to isolate y, giving us y equals 13 minus 3x, all divided by 2. This expression for y will be substituted into the circle equation.
Now we substitute y equals 13 minus 3x over 2 into the circle equation. Starting with x squared plus y squared equals 25, we replace y with our expression. This gives us x squared plus 13 minus 3x over 2, all squared, equals 25. Expanding the squared term gives us 169 minus 78x plus 9x squared, all over 4. Multiplying through by 4 to clear the fraction, we get 4x squared plus 169 minus 78x plus 9x squared equals 100. Combining like terms and rearranging gives us the quadratic equation 13x squared minus 78x plus 69 equals zero.
We solve the quadratic equation 13x squared minus 78x plus 69 equals zero using the quadratic formula. With a equals 13, b equals negative 78, and c equals 69, we calculate the discriminant: 78 squared minus 4 times 13 times 69, which equals 6084 minus 3588, giving us 2496. The square root of 2496 equals 8 times the square root of 39. Therefore, x equals 78 plus or minus 8 root 39, all over 26, which simplifies to 39 plus or minus 4 root 39, all over 13.
Now we find the corresponding y values. For x1 equals 39 plus 4 root 39 over 13, we get y1 equals 26 minus 6 root 39 over 13. For x2 equals 39 minus 4 root 39 over 13, we get y2 equals 26 plus 6 root 39 over 13. Therefore, our two solutions are: first solution is 39 plus 4 root 39 over 13, and 26 minus 6 root 39 over 13. Second solution is 39 minus 4 root 39 over 13, and 26 plus 6 root 39 over 13. Both solutions satisfy both original equations.